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June wishes to plan for her retirement and decides to invest 1000 dollars a year into an RRSP starting age 18 until age 30. After 30, she no longer contributes any money but leaves in her RRSP until age 65. Jas on the other hand, contributes 1500 dollars a year from age 35 to age 65 in the same RRSP investment. Assuming that the interest rate was fixed at 8.5% compounded annually,

How much would each have at age 65?

Apparently the answers are, for June, 339,758.97, and for Jas, 186.322.09

Unfortunately I only know the formula if there was no annual contribution. For example, I know that once June is 30, it's a geometric series, but I'm not sure how to get the amount up to when she's 30.

2 Answers2

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Denote the balance in each account when each person is age $n$ by $a_n, b _n$ for June and Jas respectively. We have that $$a_n=\begin{cases}0&n=18\\1.085a_{n-1}+1000&n\in[19,30]\\1.085a_{n-1}&n\in[31,65]\end{cases}$$ $$b_n=\begin{cases}0&n=35\\1.085a_{n-1}+1500&n\in[19,30]\end{cases}$$ These are first order linear recurrence relations (some theory supplied here) which can be solved to give $$a_n=\begin{cases}0&n=18\\\frac{1000}{17}\left(217\left(1.085^{n-19}\right)-200\right)&n\in[19,30]\\1.085^{n-30}a_{30}&n\in[31,65]\end{cases}$$ $$b_n=\begin{cases}0&n=35\\\frac{1500}{17}\left(217\left(1.085^{n-36}\right)-200\right)&n\in[36,65]\end{cases}$$ We can then use these results to get both account balances when their ages are $65$ $$a_{65}=1.085^{65-30}a_{30}=1.085^{35}\times\frac{1000}{17}\left(217\left(1.085^{30-19}\right)-200\right)=\$339758.97$$ $$b_{65}=\frac{1500}{17}\left(217\left(1.085^{65-36}\right)-200\right)=\$186322.09$$ rounded to the nearest cent.

Peter Foreman
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The future value of a contribution $c$ compounded for time $t$ is $c(1.085)^t$

If there are multiple contributions then we sum the future value of each

$c_1(1.085)^{t_1} + c_2(1.085)^{t_2} + \cdots$

June makes 12 contributions that compound for $47, 46, \cdots, 35$ years respectively.

$\sum_\limits{n=35}^{47} 1000 (1.085)^n$

Jas makes 30 contributions that comound from $30$ to $1$ year

$\sum_\limits{n=35}^{47} 1500 (1.085)^n$

Now we use some trickery for the sum of a series.

$\sum_\limits{n=a}^{b} r^n = \frac {r^{b+1}- r^a}{1-r}$

June's account is worth $1000\frac {(1.085)^{48} - (1.085)^{35}}{0.085}$ Jas' account is worth $1500\frac {(1.085)^{31} - (1.085)}{0.085}$

But these results are slightly different from what you show as the answers.

Reverse engineering it:

$1000\frac {(1.085)^{47} - (1.085)^{35}}{0.085} = 339,789\\ 1500\frac {(1.085)^{30} - (1.085)^{0}}{0.085} = 186,322$

These would suggest that the account is evaluated when June and Jas are 64, and that June did not contribute when she was 30.

user317176
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