Finding sum of none arithmetic series

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I have a question to find the sum of the following sum:
$$
S = small1*1+2*3+3*5+4*7+...+100*199
$$

I figured out that for each element in this series the following holds:
$$
a_n = a_n-1 + 4n - 3
$$

But I don't know where to go from here, I tried subtracting some other series but that did not work very well










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    4












    $begingroup$


    I have a question to find the sum of the following sum:
    $$
    S = small1*1+2*3+3*5+4*7+...+100*199
    $$

    I figured out that for each element in this series the following holds:
    $$
    a_n = a_n-1 + 4n - 3
    $$

    But I don't know where to go from here, I tried subtracting some other series but that did not work very well










    share|cite|improve this question











    $endgroup$














      4












      4








      4


      0



      $begingroup$


      I have a question to find the sum of the following sum:
      $$
      S = small1*1+2*3+3*5+4*7+...+100*199
      $$

      I figured out that for each element in this series the following holds:
      $$
      a_n = a_n-1 + 4n - 3
      $$

      But I don't know where to go from here, I tried subtracting some other series but that did not work very well










      share|cite|improve this question











      $endgroup$




      I have a question to find the sum of the following sum:
      $$
      S = small1*1+2*3+3*5+4*7+...+100*199
      $$

      I figured out that for each element in this series the following holds:
      $$
      a_n = a_n-1 + 4n - 3
      $$

      But I don't know where to go from here, I tried subtracting some other series but that did not work very well







      sequences-and-series summation telescopic-series






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      edited Jan 26 at 15:27









      Michael Rozenberg

      103k1891195




      103k1891195










      asked Jan 26 at 14:21









      Guysudai1Guysudai1

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          Each term in the equation is $n(2n-1)$, so $$S=sum_n=1^100n(2n-1)=2sum_n=1^100 n^2-sum_n=1^100 n=2fracm(m+1)(2m+1)6-fracm(m+1)2$$
          where $m=100$.






          share|cite|improve this answer











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            3












            $begingroup$

            By the telescoping sum we obtain: $$sum_n=1^100n(2n-1)=sum_n=1^100left(fracn(n+1)(4n-1)6-frac(n-1)n(4n-5)6right)=$$
            $$=frac100cdot101cdot3996=671650.$$






            share|cite|improve this answer









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            • $begingroup$
              Those two expressions whose difference is the term to be added seems like a magic trick. How did you (and how could someone else) come up with those expressions?
              $endgroup$
              – Rory Daulton
              Jan 26 at 22:04










            • $begingroup$
              @Rory Daulton We know that the sum is divided by $n(n+1)$ and it's third degree. Thus, for we can write $sumlimits_k=1^n(n(n+1)(a(n+1)+b)-(n-1)n(an+b))$ and to find values of $a$ and $b$.
              $endgroup$
              – Michael Rozenberg
              Jan 26 at 22:36


















            2












            $begingroup$

            This is, for $n=100$,
            $s(n)
            =sum_k=1^n k(2k-1)
            =sum_k=1^n (2k^2-k)
            =2sum_k=1^n k^2
            -sum_k=1^n k
            $
            .



            Plug in the formulas for the sums
            and you are done.






            share|cite|improve this answer









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              2












              $begingroup$

              $a_n=sum_r=1^n(4r-3)+a_0=dfrac n2(1+4n-3)+a_0=2n^2-n+a_0$



              $$sum_n=1^ma_n=2sum_n=1^mn^2-sum_n=1^mn+a_0sum_n=1^m1$$



              Here $a_0=0$



              Alternatively,



              $$a_m=b_m+a+bm+cm^2$$



              $$4n-3=a_n-a_n-1=b_n-b_n-1+b+c(2n-1)=b_n-b_n-1+2c(n)+b-c$$



              WLOG set $2c=4,b-c=-3iff c=b+3$ to find $b_n=b_n-1$



              set $a=0$ so that $b_m=a_m=0$






              share|cite|improve this answer











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              • $begingroup$
                Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                $endgroup$
                – lab bhattacharjee
                Jan 26 at 14:27


















              2












              $begingroup$

              Via generating functions we first find the generating function for each term and then sum them up. I'll build from the ground up. It may look overcomplicated but it also doesn't require remembering too many special identities. Our goal is to find $S_99$ where



              beginalign*
              S_m = sum_n=0^m a_n
              endalign*



              where



              beginalign*
              a_0 &= 1 \
              a_n &= (n + 1)(2n + 1) \
              &= a_n-1 + 4n + 1
              endalign*



              Now to find the generating function for our $a_n$ we find



              beginalign*
              A(x) &= sum_n=0^infty a_n x^n \
              A(x) - a(0) &= sum_n=1^infty (a_n-1 + 4n + 1) x^n \
              A(x) - 1 &= sum_n=1^infty a_n-1x^n + sum_n=1^infty (4n +1) x^n \
              A(x) - 1 &= xsum_n=0^infty a_nx^n + (sum_n=0^infty (4n +1) x^n) - (4*0 + 1) \
              A(x) &= xA(x) + (4fracx(x - 1)^2 + frac1x-1) \
              (1 - x)A(x) &= frac4x + (1 - x)(1 - x)^2 \
              A(x) &= frac3x + 1(1 - x)^3
              endalign*



              Summing them all up is easy enough, since $S_0 = 0$:



              beginalign*
              S(x) &= sum_n=0^infty S_n x^n \
              S(x) - S(0) &= sum_n=1^infty (S_n - 1 + [y^n]A(y)) x^n \
              (1 - x)S(x) &= A(x) \
              S(x) &= frac3x + 1(1 - x)^4
              endalign*



              Now we want to find the coefficient for $S(x)$
              beginalign*
              [x^n]S(x) &= [x^n]frac3x + 1(1 - x)^4 \
              &= [x^n]left(3x + 1right) sum_n=0^inftybinomn + 33x^n tag1 \
              &= left(3[x^n-1] + [x^n]right) sum_n=0^inftybinomn + 33x^n tag2\
              &= 3binomn+23 + binomn+33\
              &= frac12(n+2)(n+1)n + frac16(n+3)(n+2)(n+1)
              endalign*



              • In (1) we use the binomial series representation


              • In (2) we use the linearity of the coefficient of operator and $[x^n]x^kS(x)=[x^n-k]S(x)$.


              Finally, plugging in $n = 99$: $frac12*101*100*99 + frac16*102*101*100 = 671 650$






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                Without using any actual intelligence, you can guess that since the terms are roughly quadratic, the sum should be roughly cubic, so you guess that it is a third degree polynomial $an^3 + bn^2 + cn + d$. Then you plug in the values for n = 0, 1, 2, and 3, get for linear equations with four unknowns, and solve to get a, b, c and d.



                Try n = 4, 5, 6 to make sure this actually worked, and then it would be easy to prove by induction.






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                  6 Answers
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                  $begingroup$

                  Each term in the equation is $n(2n-1)$, so $$S=sum_n=1^100n(2n-1)=2sum_n=1^100 n^2-sum_n=1^100 n=2fracm(m+1)(2m+1)6-fracm(m+1)2$$
                  where $m=100$.






                  share|cite|improve this answer











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                    $begingroup$

                    Each term in the equation is $n(2n-1)$, so $$S=sum_n=1^100n(2n-1)=2sum_n=1^100 n^2-sum_n=1^100 n=2fracm(m+1)(2m+1)6-fracm(m+1)2$$
                    where $m=100$.






                    share|cite|improve this answer











                    $endgroup$















                      3












                      3








                      3





                      $begingroup$

                      Each term in the equation is $n(2n-1)$, so $$S=sum_n=1^100n(2n-1)=2sum_n=1^100 n^2-sum_n=1^100 n=2fracm(m+1)(2m+1)6-fracm(m+1)2$$
                      where $m=100$.






                      share|cite|improve this answer











                      $endgroup$



                      Each term in the equation is $n(2n-1)$, so $$S=sum_n=1^100n(2n-1)=2sum_n=1^100 n^2-sum_n=1^100 n=2fracm(m+1)(2m+1)6-fracm(m+1)2$$
                      where $m=100$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 26 at 18:20

























                      answered Jan 26 at 14:24









                      gunesgunes

                      3947




                      3947





















                          3












                          $begingroup$

                          By the telescoping sum we obtain: $$sum_n=1^100n(2n-1)=sum_n=1^100left(fracn(n+1)(4n-1)6-frac(n-1)n(4n-5)6right)=$$
                          $$=frac100cdot101cdot3996=671650.$$






                          share|cite|improve this answer









                          $endgroup$












                          • $begingroup$
                            Those two expressions whose difference is the term to be added seems like a magic trick. How did you (and how could someone else) come up with those expressions?
                            $endgroup$
                            – Rory Daulton
                            Jan 26 at 22:04










                          • $begingroup$
                            @Rory Daulton We know that the sum is divided by $n(n+1)$ and it's third degree. Thus, for we can write $sumlimits_k=1^n(n(n+1)(a(n+1)+b)-(n-1)n(an+b))$ and to find values of $a$ and $b$.
                            $endgroup$
                            – Michael Rozenberg
                            Jan 26 at 22:36















                          3












                          $begingroup$

                          By the telescoping sum we obtain: $$sum_n=1^100n(2n-1)=sum_n=1^100left(fracn(n+1)(4n-1)6-frac(n-1)n(4n-5)6right)=$$
                          $$=frac100cdot101cdot3996=671650.$$






                          share|cite|improve this answer









                          $endgroup$












                          • $begingroup$
                            Those two expressions whose difference is the term to be added seems like a magic trick. How did you (and how could someone else) come up with those expressions?
                            $endgroup$
                            – Rory Daulton
                            Jan 26 at 22:04










                          • $begingroup$
                            @Rory Daulton We know that the sum is divided by $n(n+1)$ and it's third degree. Thus, for we can write $sumlimits_k=1^n(n(n+1)(a(n+1)+b)-(n-1)n(an+b))$ and to find values of $a$ and $b$.
                            $endgroup$
                            – Michael Rozenberg
                            Jan 26 at 22:36













                          3












                          3








                          3





                          $begingroup$

                          By the telescoping sum we obtain: $$sum_n=1^100n(2n-1)=sum_n=1^100left(fracn(n+1)(4n-1)6-frac(n-1)n(4n-5)6right)=$$
                          $$=frac100cdot101cdot3996=671650.$$






                          share|cite|improve this answer









                          $endgroup$



                          By the telescoping sum we obtain: $$sum_n=1^100n(2n-1)=sum_n=1^100left(fracn(n+1)(4n-1)6-frac(n-1)n(4n-5)6right)=$$
                          $$=frac100cdot101cdot3996=671650.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 26 at 15:15









                          Michael RozenbergMichael Rozenberg

                          103k1891195




                          103k1891195











                          • $begingroup$
                            Those two expressions whose difference is the term to be added seems like a magic trick. How did you (and how could someone else) come up with those expressions?
                            $endgroup$
                            – Rory Daulton
                            Jan 26 at 22:04










                          • $begingroup$
                            @Rory Daulton We know that the sum is divided by $n(n+1)$ and it's third degree. Thus, for we can write $sumlimits_k=1^n(n(n+1)(a(n+1)+b)-(n-1)n(an+b))$ and to find values of $a$ and $b$.
                            $endgroup$
                            – Michael Rozenberg
                            Jan 26 at 22:36
















                          • $begingroup$
                            Those two expressions whose difference is the term to be added seems like a magic trick. How did you (and how could someone else) come up with those expressions?
                            $endgroup$
                            – Rory Daulton
                            Jan 26 at 22:04










                          • $begingroup$
                            @Rory Daulton We know that the sum is divided by $n(n+1)$ and it's third degree. Thus, for we can write $sumlimits_k=1^n(n(n+1)(a(n+1)+b)-(n-1)n(an+b))$ and to find values of $a$ and $b$.
                            $endgroup$
                            – Michael Rozenberg
                            Jan 26 at 22:36















                          $begingroup$
                          Those two expressions whose difference is the term to be added seems like a magic trick. How did you (and how could someone else) come up with those expressions?
                          $endgroup$
                          – Rory Daulton
                          Jan 26 at 22:04




                          $begingroup$
                          Those two expressions whose difference is the term to be added seems like a magic trick. How did you (and how could someone else) come up with those expressions?
                          $endgroup$
                          – Rory Daulton
                          Jan 26 at 22:04












                          $begingroup$
                          @Rory Daulton We know that the sum is divided by $n(n+1)$ and it's third degree. Thus, for we can write $sumlimits_k=1^n(n(n+1)(a(n+1)+b)-(n-1)n(an+b))$ and to find values of $a$ and $b$.
                          $endgroup$
                          – Michael Rozenberg
                          Jan 26 at 22:36




                          $begingroup$
                          @Rory Daulton We know that the sum is divided by $n(n+1)$ and it's third degree. Thus, for we can write $sumlimits_k=1^n(n(n+1)(a(n+1)+b)-(n-1)n(an+b))$ and to find values of $a$ and $b$.
                          $endgroup$
                          – Michael Rozenberg
                          Jan 26 at 22:36











                          2












                          $begingroup$

                          This is, for $n=100$,
                          $s(n)
                          =sum_k=1^n k(2k-1)
                          =sum_k=1^n (2k^2-k)
                          =2sum_k=1^n k^2
                          -sum_k=1^n k
                          $
                          .



                          Plug in the formulas for the sums
                          and you are done.






                          share|cite|improve this answer









                          $endgroup$

















                            2












                            $begingroup$

                            This is, for $n=100$,
                            $s(n)
                            =sum_k=1^n k(2k-1)
                            =sum_k=1^n (2k^2-k)
                            =2sum_k=1^n k^2
                            -sum_k=1^n k
                            $
                            .



                            Plug in the formulas for the sums
                            and you are done.






                            share|cite|improve this answer









                            $endgroup$















                              2












                              2








                              2





                              $begingroup$

                              This is, for $n=100$,
                              $s(n)
                              =sum_k=1^n k(2k-1)
                              =sum_k=1^n (2k^2-k)
                              =2sum_k=1^n k^2
                              -sum_k=1^n k
                              $
                              .



                              Plug in the formulas for the sums
                              and you are done.






                              share|cite|improve this answer









                              $endgroup$



                              This is, for $n=100$,
                              $s(n)
                              =sum_k=1^n k(2k-1)
                              =sum_k=1^n (2k^2-k)
                              =2sum_k=1^n k^2
                              -sum_k=1^n k
                              $
                              .



                              Plug in the formulas for the sums
                              and you are done.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 26 at 14:31









                              marty cohenmarty cohen

                              73.7k549128




                              73.7k549128





















                                  2












                                  $begingroup$

                                  $a_n=sum_r=1^n(4r-3)+a_0=dfrac n2(1+4n-3)+a_0=2n^2-n+a_0$



                                  $$sum_n=1^ma_n=2sum_n=1^mn^2-sum_n=1^mn+a_0sum_n=1^m1$$



                                  Here $a_0=0$



                                  Alternatively,



                                  $$a_m=b_m+a+bm+cm^2$$



                                  $$4n-3=a_n-a_n-1=b_n-b_n-1+b+c(2n-1)=b_n-b_n-1+2c(n)+b-c$$



                                  WLOG set $2c=4,b-c=-3iff c=b+3$ to find $b_n=b_n-1$



                                  set $a=0$ so that $b_m=a_m=0$






                                  share|cite|improve this answer











                                  $endgroup$












                                  • $begingroup$
                                    Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                                    $endgroup$
                                    – lab bhattacharjee
                                    Jan 26 at 14:27















                                  2












                                  $begingroup$

                                  $a_n=sum_r=1^n(4r-3)+a_0=dfrac n2(1+4n-3)+a_0=2n^2-n+a_0$



                                  $$sum_n=1^ma_n=2sum_n=1^mn^2-sum_n=1^mn+a_0sum_n=1^m1$$



                                  Here $a_0=0$



                                  Alternatively,



                                  $$a_m=b_m+a+bm+cm^2$$



                                  $$4n-3=a_n-a_n-1=b_n-b_n-1+b+c(2n-1)=b_n-b_n-1+2c(n)+b-c$$



                                  WLOG set $2c=4,b-c=-3iff c=b+3$ to find $b_n=b_n-1$



                                  set $a=0$ so that $b_m=a_m=0$






                                  share|cite|improve this answer











                                  $endgroup$












                                  • $begingroup$
                                    Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                                    $endgroup$
                                    – lab bhattacharjee
                                    Jan 26 at 14:27













                                  2












                                  2








                                  2





                                  $begingroup$

                                  $a_n=sum_r=1^n(4r-3)+a_0=dfrac n2(1+4n-3)+a_0=2n^2-n+a_0$



                                  $$sum_n=1^ma_n=2sum_n=1^mn^2-sum_n=1^mn+a_0sum_n=1^m1$$



                                  Here $a_0=0$



                                  Alternatively,



                                  $$a_m=b_m+a+bm+cm^2$$



                                  $$4n-3=a_n-a_n-1=b_n-b_n-1+b+c(2n-1)=b_n-b_n-1+2c(n)+b-c$$



                                  WLOG set $2c=4,b-c=-3iff c=b+3$ to find $b_n=b_n-1$



                                  set $a=0$ so that $b_m=a_m=0$






                                  share|cite|improve this answer











                                  $endgroup$



                                  $a_n=sum_r=1^n(4r-3)+a_0=dfrac n2(1+4n-3)+a_0=2n^2-n+a_0$



                                  $$sum_n=1^ma_n=2sum_n=1^mn^2-sum_n=1^mn+a_0sum_n=1^m1$$



                                  Here $a_0=0$



                                  Alternatively,



                                  $$a_m=b_m+a+bm+cm^2$$



                                  $$4n-3=a_n-a_n-1=b_n-b_n-1+b+c(2n-1)=b_n-b_n-1+2c(n)+b-c$$



                                  WLOG set $2c=4,b-c=-3iff c=b+3$ to find $b_n=b_n-1$



                                  set $a=0$ so that $b_m=a_m=0$







                                  share|cite|improve this answer














                                  share|cite|improve this answer



                                  share|cite|improve this answer








                                  edited Jan 26 at 14:31

























                                  answered Jan 26 at 14:25









                                  lab bhattacharjeelab bhattacharjee

                                  225k15157275




                                  225k15157275











                                  • $begingroup$
                                    Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                                    $endgroup$
                                    – lab bhattacharjee
                                    Jan 26 at 14:27
















                                  • $begingroup$
                                    Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                                    $endgroup$
                                    – lab bhattacharjee
                                    Jan 26 at 14:27















                                  $begingroup$
                                  Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                                  $endgroup$
                                  – lab bhattacharjee
                                  Jan 26 at 14:27




                                  $begingroup$
                                  Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                                  $endgroup$
                                  – lab bhattacharjee
                                  Jan 26 at 14:27











                                  2












                                  $begingroup$

                                  Via generating functions we first find the generating function for each term and then sum them up. I'll build from the ground up. It may look overcomplicated but it also doesn't require remembering too many special identities. Our goal is to find $S_99$ where



                                  beginalign*
                                  S_m = sum_n=0^m a_n
                                  endalign*



                                  where



                                  beginalign*
                                  a_0 &= 1 \
                                  a_n &= (n + 1)(2n + 1) \
                                  &= a_n-1 + 4n + 1
                                  endalign*



                                  Now to find the generating function for our $a_n$ we find



                                  beginalign*
                                  A(x) &= sum_n=0^infty a_n x^n \
                                  A(x) - a(0) &= sum_n=1^infty (a_n-1 + 4n + 1) x^n \
                                  A(x) - 1 &= sum_n=1^infty a_n-1x^n + sum_n=1^infty (4n +1) x^n \
                                  A(x) - 1 &= xsum_n=0^infty a_nx^n + (sum_n=0^infty (4n +1) x^n) - (4*0 + 1) \
                                  A(x) &= xA(x) + (4fracx(x - 1)^2 + frac1x-1) \
                                  (1 - x)A(x) &= frac4x + (1 - x)(1 - x)^2 \
                                  A(x) &= frac3x + 1(1 - x)^3
                                  endalign*



                                  Summing them all up is easy enough, since $S_0 = 0$:



                                  beginalign*
                                  S(x) &= sum_n=0^infty S_n x^n \
                                  S(x) - S(0) &= sum_n=1^infty (S_n - 1 + [y^n]A(y)) x^n \
                                  (1 - x)S(x) &= A(x) \
                                  S(x) &= frac3x + 1(1 - x)^4
                                  endalign*



                                  Now we want to find the coefficient for $S(x)$
                                  beginalign*
                                  [x^n]S(x) &= [x^n]frac3x + 1(1 - x)^4 \
                                  &= [x^n]left(3x + 1right) sum_n=0^inftybinomn + 33x^n tag1 \
                                  &= left(3[x^n-1] + [x^n]right) sum_n=0^inftybinomn + 33x^n tag2\
                                  &= 3binomn+23 + binomn+33\
                                  &= frac12(n+2)(n+1)n + frac16(n+3)(n+2)(n+1)
                                  endalign*



                                  • In (1) we use the binomial series representation


                                  • In (2) we use the linearity of the coefficient of operator and $[x^n]x^kS(x)=[x^n-k]S(x)$.


                                  Finally, plugging in $n = 99$: $frac12*101*100*99 + frac16*102*101*100 = 671 650$






                                  share|cite|improve this answer









                                  $endgroup$

















                                    2












                                    $begingroup$

                                    Via generating functions we first find the generating function for each term and then sum them up. I'll build from the ground up. It may look overcomplicated but it also doesn't require remembering too many special identities. Our goal is to find $S_99$ where



                                    beginalign*
                                    S_m = sum_n=0^m a_n
                                    endalign*



                                    where



                                    beginalign*
                                    a_0 &= 1 \
                                    a_n &= (n + 1)(2n + 1) \
                                    &= a_n-1 + 4n + 1
                                    endalign*



                                    Now to find the generating function for our $a_n$ we find



                                    beginalign*
                                    A(x) &= sum_n=0^infty a_n x^n \
                                    A(x) - a(0) &= sum_n=1^infty (a_n-1 + 4n + 1) x^n \
                                    A(x) - 1 &= sum_n=1^infty a_n-1x^n + sum_n=1^infty (4n +1) x^n \
                                    A(x) - 1 &= xsum_n=0^infty a_nx^n + (sum_n=0^infty (4n +1) x^n) - (4*0 + 1) \
                                    A(x) &= xA(x) + (4fracx(x - 1)^2 + frac1x-1) \
                                    (1 - x)A(x) &= frac4x + (1 - x)(1 - x)^2 \
                                    A(x) &= frac3x + 1(1 - x)^3
                                    endalign*



                                    Summing them all up is easy enough, since $S_0 = 0$:



                                    beginalign*
                                    S(x) &= sum_n=0^infty S_n x^n \
                                    S(x) - S(0) &= sum_n=1^infty (S_n - 1 + [y^n]A(y)) x^n \
                                    (1 - x)S(x) &= A(x) \
                                    S(x) &= frac3x + 1(1 - x)^4
                                    endalign*



                                    Now we want to find the coefficient for $S(x)$
                                    beginalign*
                                    [x^n]S(x) &= [x^n]frac3x + 1(1 - x)^4 \
                                    &= [x^n]left(3x + 1right) sum_n=0^inftybinomn + 33x^n tag1 \
                                    &= left(3[x^n-1] + [x^n]right) sum_n=0^inftybinomn + 33x^n tag2\
                                    &= 3binomn+23 + binomn+33\
                                    &= frac12(n+2)(n+1)n + frac16(n+3)(n+2)(n+1)
                                    endalign*



                                    • In (1) we use the binomial series representation


                                    • In (2) we use the linearity of the coefficient of operator and $[x^n]x^kS(x)=[x^n-k]S(x)$.


                                    Finally, plugging in $n = 99$: $frac12*101*100*99 + frac16*102*101*100 = 671 650$






                                    share|cite|improve this answer









                                    $endgroup$















                                      2












                                      2








                                      2





                                      $begingroup$

                                      Via generating functions we first find the generating function for each term and then sum them up. I'll build from the ground up. It may look overcomplicated but it also doesn't require remembering too many special identities. Our goal is to find $S_99$ where



                                      beginalign*
                                      S_m = sum_n=0^m a_n
                                      endalign*



                                      where



                                      beginalign*
                                      a_0 &= 1 \
                                      a_n &= (n + 1)(2n + 1) \
                                      &= a_n-1 + 4n + 1
                                      endalign*



                                      Now to find the generating function for our $a_n$ we find



                                      beginalign*
                                      A(x) &= sum_n=0^infty a_n x^n \
                                      A(x) - a(0) &= sum_n=1^infty (a_n-1 + 4n + 1) x^n \
                                      A(x) - 1 &= sum_n=1^infty a_n-1x^n + sum_n=1^infty (4n +1) x^n \
                                      A(x) - 1 &= xsum_n=0^infty a_nx^n + (sum_n=0^infty (4n +1) x^n) - (4*0 + 1) \
                                      A(x) &= xA(x) + (4fracx(x - 1)^2 + frac1x-1) \
                                      (1 - x)A(x) &= frac4x + (1 - x)(1 - x)^2 \
                                      A(x) &= frac3x + 1(1 - x)^3
                                      endalign*



                                      Summing them all up is easy enough, since $S_0 = 0$:



                                      beginalign*
                                      S(x) &= sum_n=0^infty S_n x^n \
                                      S(x) - S(0) &= sum_n=1^infty (S_n - 1 + [y^n]A(y)) x^n \
                                      (1 - x)S(x) &= A(x) \
                                      S(x) &= frac3x + 1(1 - x)^4
                                      endalign*



                                      Now we want to find the coefficient for $S(x)$
                                      beginalign*
                                      [x^n]S(x) &= [x^n]frac3x + 1(1 - x)^4 \
                                      &= [x^n]left(3x + 1right) sum_n=0^inftybinomn + 33x^n tag1 \
                                      &= left(3[x^n-1] + [x^n]right) sum_n=0^inftybinomn + 33x^n tag2\
                                      &= 3binomn+23 + binomn+33\
                                      &= frac12(n+2)(n+1)n + frac16(n+3)(n+2)(n+1)
                                      endalign*



                                      • In (1) we use the binomial series representation


                                      • In (2) we use the linearity of the coefficient of operator and $[x^n]x^kS(x)=[x^n-k]S(x)$.


                                      Finally, plugging in $n = 99$: $frac12*101*100*99 + frac16*102*101*100 = 671 650$






                                      share|cite|improve this answer









                                      $endgroup$



                                      Via generating functions we first find the generating function for each term and then sum them up. I'll build from the ground up. It may look overcomplicated but it also doesn't require remembering too many special identities. Our goal is to find $S_99$ where



                                      beginalign*
                                      S_m = sum_n=0^m a_n
                                      endalign*



                                      where



                                      beginalign*
                                      a_0 &= 1 \
                                      a_n &= (n + 1)(2n + 1) \
                                      &= a_n-1 + 4n + 1
                                      endalign*



                                      Now to find the generating function for our $a_n$ we find



                                      beginalign*
                                      A(x) &= sum_n=0^infty a_n x^n \
                                      A(x) - a(0) &= sum_n=1^infty (a_n-1 + 4n + 1) x^n \
                                      A(x) - 1 &= sum_n=1^infty a_n-1x^n + sum_n=1^infty (4n +1) x^n \
                                      A(x) - 1 &= xsum_n=0^infty a_nx^n + (sum_n=0^infty (4n +1) x^n) - (4*0 + 1) \
                                      A(x) &= xA(x) + (4fracx(x - 1)^2 + frac1x-1) \
                                      (1 - x)A(x) &= frac4x + (1 - x)(1 - x)^2 \
                                      A(x) &= frac3x + 1(1 - x)^3
                                      endalign*



                                      Summing them all up is easy enough, since $S_0 = 0$:



                                      beginalign*
                                      S(x) &= sum_n=0^infty S_n x^n \
                                      S(x) - S(0) &= sum_n=1^infty (S_n - 1 + [y^n]A(y)) x^n \
                                      (1 - x)S(x) &= A(x) \
                                      S(x) &= frac3x + 1(1 - x)^4
                                      endalign*



                                      Now we want to find the coefficient for $S(x)$
                                      beginalign*
                                      [x^n]S(x) &= [x^n]frac3x + 1(1 - x)^4 \
                                      &= [x^n]left(3x + 1right) sum_n=0^inftybinomn + 33x^n tag1 \
                                      &= left(3[x^n-1] + [x^n]right) sum_n=0^inftybinomn + 33x^n tag2\
                                      &= 3binomn+23 + binomn+33\
                                      &= frac12(n+2)(n+1)n + frac16(n+3)(n+2)(n+1)
                                      endalign*



                                      • In (1) we use the binomial series representation


                                      • In (2) we use the linearity of the coefficient of operator and $[x^n]x^kS(x)=[x^n-k]S(x)$.


                                      Finally, plugging in $n = 99$: $frac12*101*100*99 + frac16*102*101*100 = 671 650$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 26 at 20:07









                                      WorldSEnderWorldSEnder

                                      440313




                                      440313





















                                          2












                                          $begingroup$

                                          Without using any actual intelligence, you can guess that since the terms are roughly quadratic, the sum should be roughly cubic, so you guess that it is a third degree polynomial $an^3 + bn^2 + cn + d$. Then you plug in the values for n = 0, 1, 2, and 3, get for linear equations with four unknowns, and solve to get a, b, c and d.



                                          Try n = 4, 5, 6 to make sure this actually worked, and then it would be easy to prove by induction.






                                          share|cite|improve this answer









                                          $endgroup$

















                                            2












                                            $begingroup$

                                            Without using any actual intelligence, you can guess that since the terms are roughly quadratic, the sum should be roughly cubic, so you guess that it is a third degree polynomial $an^3 + bn^2 + cn + d$. Then you plug in the values for n = 0, 1, 2, and 3, get for linear equations with four unknowns, and solve to get a, b, c and d.



                                            Try n = 4, 5, 6 to make sure this actually worked, and then it would be easy to prove by induction.






                                            share|cite|improve this answer









                                            $endgroup$















                                              2












                                              2








                                              2





                                              $begingroup$

                                              Without using any actual intelligence, you can guess that since the terms are roughly quadratic, the sum should be roughly cubic, so you guess that it is a third degree polynomial $an^3 + bn^2 + cn + d$. Then you plug in the values for n = 0, 1, 2, and 3, get for linear equations with four unknowns, and solve to get a, b, c and d.



                                              Try n = 4, 5, 6 to make sure this actually worked, and then it would be easy to prove by induction.






                                              share|cite|improve this answer









                                              $endgroup$



                                              Without using any actual intelligence, you can guess that since the terms are roughly quadratic, the sum should be roughly cubic, so you guess that it is a third degree polynomial $an^3 + bn^2 + cn + d$. Then you plug in the values for n = 0, 1, 2, and 3, get for linear equations with four unknowns, and solve to get a, b, c and d.



                                              Try n = 4, 5, 6 to make sure this actually worked, and then it would be easy to prove by induction.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Jan 26 at 20:16









                                              gnasher729gnasher729

                                              6,0151028




                                              6,0151028



























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