You plan to retire in 30 years and decide to save $10,000 per year. If the interest rate is 6% compounded monthly, how much will you have in 30 years? Assume that each deposit is made at the end of every year.

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Woodland Hills, CA

New Wilmington, PA

This problem asks for the future value of an annuity with monthly deposits.

The standard formula for this amount is

FV =R*((1+i)^{n}-1)/i,

where R is the payment per compounding period, i = the interest rate per compounding period, and n the number of compounding periods. This formula assumes that one payment is made each compounding period, which in your problem is not the case: the compounding period is a month (n=30*12), while a payment R=10,000 is made once year.

We can get an approximation if we assume instead that a payment of R=$10,000/12 is made every month.

With i=0.06/12 we get

FV =(10000/12) *((1+(0.06/12))^{30*12}-1)/(0.06/12)

=$ 837,100

This is an *over*estimate, because it assumes monthly payments with monthly compounding.

We can get an *under*estimate if we assume annual compounding with annual payments. In this case n=30, i=0.06 and

FV=10000 * ((1+0.06)^{30}-1)/0.06

=$ 790,580

The actual answer will lie between these two estimates.

Willowbrook, IL

Since the first deposit will be made by the end of the first year, there will be 29 deposits.

The first deposit will grow to be 10,000*(1+0.06/12)^{12*29}; the second deposit grows to be 10,000*(1+0.06/12)^{12*28} and so on. The last deposit will grow to 10,000*(1+0.06/12)^{12}≈$10616.78.

We need to find the sum 10,000*(1+0.06/12)^{12}+10,000*(1+0.06/12)^{12*2}+…+10,000*(1+0.06/12)^{12*29}=

=10000*[(1+0.06/12)^{12}+((1+0.06/12)^{12})^{2}+…+((1+0.06/12)^{12})^{29}]

So we need to sum geometrical progression with the first term (1+0.06/12)^{12}≈1.061678 and the base 1.061678, the total number of terms being 29.

S=q*(q^{N-1}-1)/(q-1), where q is the base of geometrical progression, N is the number of terms.

Plug in the number to get:

S≈80.432448

Final answer is S*10000=804324.48

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