Question

A bank gives 3% interest on its savings accounts, compounded monthly. If one opens a savings account and deposits $ 300 each month for a total of 11 years, how much money will be in the account after the last deposit is made?

Answer 1

3% is spread over 12 months; and the interest as calculated and added each month

Answer 2

B[n] = B_[n-1] + B{n-1}(.03/12)

B[n] = B_[n-1] (1+(.03/12))

B[n] = B[0](1+(.03/12))^n

where n is measured in months

Answer 3

gotta add in the 300 per month tho ... doh!!

Answer 4

B[n] = B_[n-1] + B{n-1}(.03/12)

B[n] = B_[n-1] (1+(.03/12)) + 300n

hmmm....

Answer 5

B[n] = B_[n-1] (1+(.03/12)) + 300n

B[n-1] = B_[n-2] (1+(.03/12)) + 300(n-1)

B[n] = (B_[n-2] (1+(.03/12)) + 300(n-1)) (1+(.03/12)) + 300n


B[n] = B_[n-2] (401/400))^2 + (401/400)300n- (401/400)300 + 300n

B[n] = B_[n-2] (401/400))^2 + 300(401/400)n- (401/400) + n)

\[B_n = B_{n-2} (401/400)^2 + 300(\frac{401}{400}n-\frac{401}{400} + n)\]

oy, this would work out better on paper, and also if i knew an easier trick :)