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You have an initial amount of money saved $P$ (for principal). On this, you get an annual rate of return on an investment equal to $r$ yearly return on your investment). Assuming you don't withdraw from your savings or save any extra, the amount of money you have, $a_n$ in year $n$ is the amount you had invested in year $n - 1$ plus the interest accrued. We use the convention that $n = 0$ corresponds to the year of the initial investment, so $a_0 =P$.

I would like to know how to get the recurrence relation and find a formula to solve for $a_n$.

mdave16
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Jeje
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1 Answers1

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Using the information you gave, there are 2 key steps to do this.

Interest

So if we have $£x$, and we receive interest at a rate of $r$, we get $£x$ (our initial investment) + $£rx$ (the interest) (here we view the investment as a decimal). So for example, if my bank was giving me $20\%$ interest, and I invest $£20$, then in total I would have $£20 + £20\times0.2 = £24$.

An easier way to put this would be $£x(1+r)$.

Recurrences

We have that $a_n$ is the money from the $n-1$- th year along with interest, from above, we have that $$a_n = a_{n-1}(1+r).$$ Along with the initial condition $a_0 = P$, we can solve this.

One general way to solve recurrence relations, is to write them out until we see a pattern emerge.

So \begin{align} a_n &= a_{n-1}(1+r)\\ &= a_{n-2}(1+r)^2\\ &\,\vdots\\ &= a_{1}(1+r)^{n-1}\\ &= a_0(1+r)^n\\ &= P(1+r)^n. \end{align}

Thus we have solved the recurrence!

Recurrences with additional investment

What if I invested an additional $P$ each year? What is the recurrence equation and the formula? Is it $a_n = n P(1+r) + P$?

Not quite, let's apply the same technique. Each year, we would get interest and then also we would add in investment. This makes it $$a_n = a_{n-1}(1+r) + P.$$

By applying the same technique as above, we get that $$a_n = P \frac{(1+r)^{n+1} - 1}{r}.$$

mdave16
  • 951