Hi Steph,

At the end of the first year the height of the tree is

\[86 + 42 \mbox{ cm}\]

at the end of the second year its height is

\[86 + 42 + 0.95 \times 42 \mbox{ cm.}\]

At the end of the third its height is

\[86 + 42 + 42 \times 0.95 + 42 \times 0.95 \times 0.95 \mbox{ cm}\]

or

\[86 + 42 + 42 \times 0.95 + 42 \times 0.95^2 \mbox{ cm}\]

and so on.

The pattern is 86 plus the geometric series $a + ar + ar^2 + ar^3 \cdot \cdot \cdot$ where $a = 42$ and $r = 0.95.$

There are formulae for the sum of $n$ terms of a geometric series and for the sum of an infinite number of terms but it is much more useful to remember how to generate the expressions than to memorize them.

Suppose that the sum of the first $n$ terms is $S_n,$ then

\[S_n = a + ar + ar^2 + ar^3 \cdot \cdot \cdot + ar^{n-1}.\]

multiply each side of the equation by $r$ to obtain

\[S_nr = ar + ar^2 + ar^3 \cdot \cdot \cdot + ar^{n-1} + ar^n.\]

Subtract the second equation from the first and you get

\[S_n - S_nr = S_n(1 - r) = a - ar^n = a(1 - r^n).\]

Hence

\[S_n = \frac{a(1-r^n)}{1 - r}.\]

You can use this expression to calculate the height of the tree after 5 years.

In your problem $r = 0.95$ which is less than 1 and hence as $n$ becomes large, $r^n$ approaches zero and $S_n$ approaches $\large \frac{a}{1 - r}.$

Penny