Sum of Geometric progression?
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A man deposits $200 at the beginning of every year into a bank account at a compund interest rate of 3%per annum. Find out how much he has at the end of 10th year to the nearest dollar?
I honestly don't quite understand the part about "depositing $200 every year".
How do you go about solving this?
sequences-and-series geometric-progressions
asked 4 hours ago
Henias
214
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up vote
1
down vote
favorite
A man deposits $200 at the beginning of every year into a bank account at a compund interest rate of 3%per annum. Find out how much he has at the end of 10th year to the nearest dollar?
I honestly don't quite understand the part about "depositing $200 every year".
How do you go about solving this?
sequences-and-series geometric-progressions
asked 4 hours ago
Henias
214
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A man deposits $200 at the beginning of every year into a bank account at a compund interest rate of 3%per annum. Find out how much he has at the end of 10th year to the nearest dollar?
I honestly don't quite understand the part about "depositing $200 every year".
How do you go about solving this?
sequences-and-series geometric-progressions
asked 4 hours ago
Henias
214
A man deposits $200 at the beginning of every year into a bank account at a compund interest rate of 3%per annum. Find out how much he has at the end of 10th year to the nearest dollar?
I honestly don't quite understand the part about "depositing $200 every year".
How do you go about solving this?
sequences-and-series geometric-progressions
sequences-and-series geometric-progressions
asked 4 hours ago
Henias
214
asked 4 hours ago
Henias
214
edited 4 hours ago
asked 4 hours ago
Henias
214
asked 4 hours ago
Henias
214
asked 4 hours ago
Henias
214
214
add a comment |Â
add a comment |Â
2 Answers
2
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4
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accepted
The first $$200$ compounds for $10$ years
the next year he deposits another $$200$ into the account, and that will compound for $9$ more years.
After 10 years he has:
$200(1.03)^10 + 200(1.03)^9 + cdots + 200(1.03)$
And how would you sum that up?
answered 4 hours ago
Doug M
41.4k31752
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up vote
3
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At the end of the first year it's going to be $200 cdot 1.03$;
at the end of the second year it's going to be $(200 cdot 1.03 + 200) cdot 1.03=200 cdot (1.03+1.03^2)$;
at the end of the third year we will have $200 cdot (1.03+1.03^2+1.03^3)$.
Do you see the pattern?
answered 4 hours ago
Vasya
3,0491514
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The first $$200$ compounds for $10$ years
the next year he deposits another $$200$ into the account, and that will compound for $9$ more years.
After 10 years he has:
$200(1.03)^10 + 200(1.03)^9 + cdots + 200(1.03)$
And how would you sum that up?
answered 4 hours ago
Doug M
41.4k31752
add a comment |Â
up vote
4
down vote
accepted
The first $$200$ compounds for $10$ years
the next year he deposits another $$200$ into the account, and that will compound for $9$ more years.
After 10 years he has:
$200(1.03)^10 + 200(1.03)^9 + cdots + 200(1.03)$
And how would you sum that up?
answered 4 hours ago
Doug M
41.4k31752
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The first $$200$ compounds for $10$ years
the next year he deposits another $$200$ into the account, and that will compound for $9$ more years.
After 10 years he has:
$200(1.03)^10 + 200(1.03)^9 + cdots + 200(1.03)$
And how would you sum that up?
answered 4 hours ago
Doug M
41.4k31752
The first $$200$ compounds for $10$ years
the next year he deposits another $$200$ into the account, and that will compound for $9$ more years.
After 10 years he has:
$200(1.03)^10 + 200(1.03)^9 + cdots + 200(1.03)$
And how would you sum that up?
answered 4 hours ago
Doug M
41.4k31752
answered 4 hours ago
Doug M
41.4k31752
answered 4 hours ago
Doug M
41.4k31752
answered 4 hours ago
Doug M
41.4k31752
41.4k31752
add a comment |Â
add a comment |Â
up vote
3
down vote
At the end of the first year it's going to be $200 cdot 1.03$;
at the end of the second year it's going to be $(200 cdot 1.03 + 200) cdot 1.03=200 cdot (1.03+1.03^2)$;
at the end of the third year we will have $200 cdot (1.03+1.03^2+1.03^3)$.
Do you see the pattern?
answered 4 hours ago
Vasya
3,0491514
add a comment |Â
up vote
3
down vote
At the end of the first year it's going to be $200 cdot 1.03$;
at the end of the second year it's going to be $(200 cdot 1.03 + 200) cdot 1.03=200 cdot (1.03+1.03^2)$;
at the end of the third year we will have $200 cdot (1.03+1.03^2+1.03^3)$.
Do you see the pattern?
answered 4 hours ago
Vasya
3,0491514
add a comment |Â
up vote
3
down vote
up vote
3
down vote
At the end of the first year it's going to be $200 cdot 1.03$;
at the end of the second year it's going to be $(200 cdot 1.03 + 200) cdot 1.03=200 cdot (1.03+1.03^2)$;
at the end of the third year we will have $200 cdot (1.03+1.03^2+1.03^3)$.
Do you see the pattern?
answered 4 hours ago
Vasya
3,0491514
At the end of the first year it's going to be $200 cdot 1.03$;
at the end of the second year it's going to be $(200 cdot 1.03 + 200) cdot 1.03=200 cdot (1.03+1.03^2)$;
at the end of the third year we will have $200 cdot (1.03+1.03^2+1.03^3)$.
Do you see the pattern?
answered 4 hours ago
Vasya
3,0491514
answered 4 hours ago
Vasya
3,0491514
answered 4 hours ago
Vasya
3,0491514
answered 4 hours ago
Vasya
3,0491514
3,0491514
add a comment |Â
add a comment |Â
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