Problem source

My bank pays $\rho\%$ interest at the end of each year. I start with nothing in my account. Then for $m$ years I deposit $\pounds a$ in my  account at the beginning of each year. After the end of the $m$th year, I neither deposit nor withdraw for $l$ years.
Show that the total amount in my account at the end of this period is \[\pounds a\frac{r^{l+1}(r^{m}-1)}{r-1}\] where $r=1+{\displaystyle \frac{\rho}{100}}$.
At the beginning of each of the  $n$ years following this period I withdraw $\pounds b$ and this leaves my account empty after the $n$th withdrawal.
Find an expression for $a/b$ in terms of $r$, $l$, $m$ and $n$.

Solution source

Rather than putting the deposits in the same account, imagine they are all put in separate accounts. Then for example, the first $\pounds a$ will go on to become $\pounds r^m \cdot r^l a$ from $m$ years of compound interest as more money is deposited, followed by $l$ years where no money is deposited.

Therefore the total amount at the end is:

\begin{align*}
&& S &= r^{m}r^l a + r^{m-1}r^l a + \cdots + r r^l a \\
&&&= r^l a(r^m + \cdots + r) \\
&&&= ar^l\frac{r^{m+1}-r}{r-1} \\
&&&= a \frac{r^{l+1}(r^m-1)}{r-1}
\end{align*}

Rather than withdrawing $b$ each time, imagine that in the $n$th year we withdraw each $b$ with the appropriate additional interest, ie

\begin{align*}
&&  \underbrace{a \frac{r^{l+1}(r^m-1)}{r-1}}_{\text{amount before $n$ years}} \underbrace{r^{n-1}}_{\text{accounting for interest}} &= b r^{n-1} + br^{n-2} + \cdots + b \\
&&&= b \frac{r^n-1}{r-1} \\
\Rightarrow && \frac{a}{b} &= \frac{r^n-1}{r^{l+n}(r^m-1)}
\end{align*}