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An investor deposits $1000 in an account paying interest at a rate of 8%, compounded monthly, and also makes additional deposits of $25 per month. Find the balance in the account after 3 years.

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Answer:

Let r be the monthly interest rate: r = .08/12, and a = 1+r

Balance after 3 years

= 1000a^36 + 25a^35 + 25a^34 + ... + 25a, assuming the last deposit earn one month interest.

= 1000a^36 + 25a(1-a^35)/(1-a)

= $2258.63 <==Answer

Attn: If the last contribution doesn't earn any interest, then the final balance should be 2258.63 + 25 = $2283.63.

Explanation:

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