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Assume that each deposit is made at the end of every year.

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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Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
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   A = P ( 1 + 0.06/12)(12*30) = 10,000*(1.005)360 = 60225.75

Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
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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 overestimate, because it assumes monthly payments with monthly compounding.
 
We can get an underestimate 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.
 
 
Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
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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*(qN-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