NO
You have to use the formula for "amount of an annuity"
i = .02/4 = .005
n = 10(4) = 40
amount = 1000( 1.005^40 - 1)/.005
= $44,158.85
What are you doing with,
Total = 1220.79
40,000 + 1220.79 = 41,220.79 ?????
An employee saves for her retirement by depositing $1000 a quarter. She deposits her money in an annuity which pays a return of 2% a year compounded quarterly. What will the value of the annuity be after ten years?
n = 4*10 = 40 interest per period (quarter)
Total = 1220.79
40,000 + 1220.79 = 41,220.79 is this correct for the value of the annuity after ten years
2 answers
1220.79 is the future value of 1000 after 10 years, which is not the value of the annuity.
The annuity formula is the sum of the future value of quarterly payments.
The first quarterly payment (always deposited at the END of the first quarter) therefore earns 39 quarters of interest, or
the future value is 1000*(1.005^39).
Similarly, the subsequent payments earn
1000*(1.005^38), 1000*(1.005^37), 1000*(1.005^36), ...., 1000(1.005^2), 1000(1.005^1), 1000(1.005^0).
The last payment is made on the due date of annuity, hence makes zero interest.
Summing them all by factorization,
Future value
= annuity
= 1000(1.005^39+1.005^38+....1.005^2+1.005^1+1.005^0)
=1000(1.005^40-1)/(1.005-1)
=1000(1.005^40-1)/0.005
=44,158.85
as Reiny gave.
[ note: ((1+i)^n-1)=(1+i-1)[(1+i)^(n-1)+(1+i)^(n-2)+(1+i)^(n-3)+...+(1+i)^(1)+(1+i)^(0)]
The annuity formula is the sum of the future value of quarterly payments.
The first quarterly payment (always deposited at the END of the first quarter) therefore earns 39 quarters of interest, or
the future value is 1000*(1.005^39).
Similarly, the subsequent payments earn
1000*(1.005^38), 1000*(1.005^37), 1000*(1.005^36), ...., 1000(1.005^2), 1000(1.005^1), 1000(1.005^0).
The last payment is made on the due date of annuity, hence makes zero interest.
Summing them all by factorization,
Future value
= annuity
= 1000(1.005^39+1.005^38+....1.005^2+1.005^1+1.005^0)
=1000(1.005^40-1)/(1.005-1)
=1000(1.005^40-1)/0.005
=44,158.85
as Reiny gave.
[ note: ((1+i)^n-1)=(1+i-1)[(1+i)^(n-1)+(1+i)^(n-2)+(1+i)^(n-3)+...+(1+i)^(1)+(1+i)^(0)]