Find the balance?

Question

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.

Answer 1

It's a messy formula: Balance(n) = P(1 + r)ⁿ + c[ ((1 + r)ⁿ - 1) / r ]

P = starting Principal = $1000
r = interest rate per period (8% / 12) = 0.00666... per month
n = the number of periods = 36
c = monthly contribution = 25

The answer comes out to be $2283.63

You can check your answer with a spreadsheet like MS Excel:

Start with balance B₀ = $1000 at month 0
B1 = B0 * (1+ .08/12) + 25 = 1031.666...
B2 = B1 * (1+ .08/12) + 25 = 1066.5444...
...
B36 = B35 * (1+ .08/12) + 25 = 2283.63

Look here for the derivation of the formula:
www.moneychimp.com/articles/finworks/fmbasinv.htm

Answer 2

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.