# Retirement Fund with Interest

A young woman 22 years of age has just graduated from college. She accepts a good job and desires to establish her own retirement fund. At the end of each year thereafter she plans to deposit 2000 in a fund at 15% annual interest. How old will she be when the fund has an accumulated value of 1000000

Using Compound Interest Formula:

F = P(1+i)^n
1000000 = 2000(1 + 0.15)^n
n = 44.465


Add to age 22+44 = 66 years old.

What am i doing wrong. Any hint?

• The question says "at the end of each year she plans to deposit 2000..". Using compound interest doesn't do that, it only takes an initial value and adds 15% to that each year.. Try writing out what happens at the end of year 1, year 2, year 3 etc. You should find it forms a geometric series. Jun 28, 2015 at 16:02

As Mattos already said the mistake is that she adds $2000$ dollars at the end of each year.

After $1$ year she will have $2000$ dollars. After two years she will have $2000\cdot1.15+2000$, After three she will have $2000\cdot1.15^2+2000\cdot1.15+2000$.

After $n+1$ years she will have $2000(1.15^n+1.15^{n-1}\dots +1)$ By formula of geometric sum this is equal to: $2000(\frac{1.15^{n+1}-1}{1.15-1})$.

So after $n$ years she will have $2000(\frac{1.15^{n}-1}{0.15}$) So we must solve:

$1,000,000\cdot0.15=2000(1.15^{n}-1)\iff 500\cdot0.15=(1.15^{n}-1)\iff 76=1.15^n$ which has an approximate solution of $n=30.9865$

Therefore she will have $22+31=53$ years

suppose she keeps depositing for $n-1$ years. then the first $2000$ earns interest for $n-1$ years, the second $2000$ earns interest for $n-2$ years and son on. so the total amount at the end of $n$ years is given by the geometric series $$2000\left(1+1.15 + {1.15}^2 + \cdots + {1.15}^{n-1})\right) = 2000\frac{{1.15}^n - 1}{.15} \to 1.15^n = 76 \to n = 30.98$$

if you add $22,$ you get approximately $53$