Savings

[Blue Bagger]

Hello again,

One last question and i will have finished all my assignments. Yay!

Anyway, the last question is:

Assuming an interest rate of 6% and a 0 starting balance over 4 years. How much needs to be deposited at the end of each MONTH to reach the value of $63123.85 at the end of the 4 years.

I am confused by my notes and have not been able to work out the correct formula to use.

Could someone please help.

Thanks.

 

[HallsofIvy]

Hello again,

One last question and i will have finished all my assignments. Yay!

Anyway, the last question is:

Assuming an interest rate of 6% and a 0 starting balance over 4 years. How much needs to be deposited at the end of each MONTH to reach the value of $63123.85 at the end of the 4 years.

I am confused by my notes and have not been able to work out the correct formula to use.

Could someone please help.

Thanks.

Look at it step by step. Originally, you have $0 but then you add "X" dollars. At 6% interest per year, in one month it will have earned (0.06/12)X= 0.005X dollars so will have increased to X+ 0.005X= 1.005X. Then you put in another X dollars so you have X+ 1.005X= (2+ 0.005)X. The interest on that is 0.005(2+ 0.005)X= (2(0.005)+ (0.005)^2)X so in the second month will have increased to (2+ 0.005)X+ (2(0.005)+ (0.005)^2)X= (2+ 3(0.005)+ (0.005)^2)X. When you deposit another X dollars you have (3+ 3(0.005)+ (0.005)^2)X. The interest on that the third month is 0.005(3+ 3(0.005)+ (0.005)^2)X= (3(0.005)+ 3(0.005)^2+ (0.005)^3)X so the total amount in the account, after the third month, is (3+ 3(0.005)+ (0.005)^2)X+ (3(0.005)+ 3(0.005)^2+ (0.005)^3)X= (3+ 4(0.005)+ 4(0.005)^2+ (0.005)^3)X.

I think that, at this point, I would conjecture that the amount in the account after the "nth" month would be (3+ (n+1)(0.005)+ (n+1)(0.005)^2+ ...+ (n+1)(0.005)^(n-1)+ (0.005)^n)X.

Can you check if that is true? If so then part of that, at least, can be written as a "geometric series" and there is a formula for its sum. Find the sum when n= 48.