# Find the maximum height that the Pohutukawa tree is expected to reach in cm?

## A Pohutukawa tree is 86 cm tall when it is planted. In the first year after it is planted, the tree grows 42cm in height. Each year the tree grows in height by 95% of the growth of the previous year.

Feb 25, 2018

Height after 5 years: 276cm

Edit
Maximum height: 926cm.

#### Explanation:

The growth of the tree over n years is

$86 + 42 \cdot {0.95}^{0} + 42 \cdot {0.95}^{1} + . . . + 42 \cdot {0.95}^{n - 1}$

$r = 0.95$
$a = 42$

The sum of a geometric progression is,

${S}_{n} = \frac{a \left(1 - {r}^{n}\right)}{1 - r}$,

Therefore the height in 5 years is 190.02cm + the initial 86cm = 276cm.

Edit I see that you have changed the question to ask about the maximum height of the tree. In this case, the formula

${S}_{n} = \frac{a}{1 - r}$ can be used, thus

$\frac{42}{1 - 0.95} = 840$

Added to the initial height of 86 cm, gives 926cm.

Feb 25, 2018

926cm

#### Explanation:

This is going to need a double check...

The tree starts at 86cm. Year one, the tree will be:

$86 c m + 42 c m$

Year two, the tree will be $86 c m + 42 c m + 42 c m \left(.95\right)$

Year three the tree will be $86 c m + 42 c m + 42 c m \left(.95\right) + 42 c m \left(.95\right) \left(.95\right)$

This goes on year after year. One of the things we can do is factor out the 42, so our tree looks like this:

$86 c m + 42 c m \left(1 + \left(.95\right) + \left(.95\right) \left(.95\right) + \ldots\right)$

All of those (.95) terms (even the 1) can be written as exponents of (.95) so:

$86 c m + 42 c m \left({\left(.95\right)}^{0} + {\left(.95\right)}^{1} + {\left(.95\right)}^{2} + \ldots + {\left(.95\right)}^{n}\right)$

If you calculate the summation of the (.95) exponential terms, you get 20

"_0^oosum.95^n=20 (Someone please check notation/math!)

Therefore, the maximum height of the tree (H) will be:

$H = 86 c m + 42 c m \left(20\right) = 926 c m$

Feb 25, 2018

$926 \text{ centimeters}$

#### Explanation:

{: ("initial height (cm):",86), ("height after 1 year:",86+(42)), ("height after 2 years:",86+( 42)+(42 * 0.95)), ("height after 3 years:",86+( 42* 0.95)+((42*0.95) * 0.95)), (,), ("height after "n" years:",86 + Sigma_(y=0)^n 42 * 0.95^y) :}

The general formula for a converging geometric series is
$\textcolor{w h i t e}{\text{XXX}} S = {\Sigma}_{i = 0}^{\infty} a i = \frac{{a}_{0}}{1 - r}$
where $r$ is the common ratio (note for convergence $\left\mid r \right\mid < 1$)
and ${a}_{i}$ is the ${i}^{\text{th}}$ term of the series (with ${a}_{0}$ being the initial value.

In this case ${a}_{0} = 42 \text{ cm.}$ and $r = 0.95$

So the final (maximum) height will be
$\frac{\textcolor{w h i t e}{\text{XXX")S=86 + (42" cm}}}{1 - 0.95}$

$\frac{\textcolor{w h i t e}{\text{XXX") =86+ (42" cm}}}{0.05}$

color(white)("XXX")=86+42" cm"xx20

color(white)("XXX")=86+840" cm"

color(white)("XXX")=926 " cm"