Geometric Series

i) a=729

 

r=\dfrac{2}{3}

 

n=7

 

Substitute into formula:

 

\begin{aligned}S_{n}&=\dfrac{a(1-r^{n})}{1-r}\\[1.2em]&=\dfrac{729\left(1-\left(\dfrac{2}{3}\right)^{7}\right)}{1-\dfrac{2}{3}}\\[1.2em]&=\dfrac{729\left(1-\dfrac{128}{2187}\right)}{\dfrac{1}{3}}\\[1.2em]&=3\times 729\left(\dfrac{2059}{2187}\right)\\[1.2em]&=2187\times\dfrac{2059}{2187}\\[1.2em]&=2059\end{aligned}

 

 

ii) \begin{aligned}S_{\infty}&=\dfrac{a}{1-r}\\[1.2em]&=\dfrac{729}{1-\dfrac{2}{3}}\\[1.2em]&=\dfrac{729}{\dfrac{1}{3}}\\[1.2em]&=3\times 729\\[1.2em]&=2187\end{aligned}

i) A sum to infinity is not sensible for two reasons:

The tree will not live forever.

r=1.01 so |r|>1 so there is no finite sum to infinity.

 

ii) a=0.2

 

r=1.01

 

n=22

 

Substitute into formula:

 

\begin{aligned}S_{n}&=\dfrac{a(1-r^{n})}{1-r}\\[1.2em]&=\dfrac{0.2(1-1.01^{21})}{1-1.01}\\[1.2em]&=4.65\end{aligned}m

S_{\infty}=\dfrac{a}{1-r}

 

a=2

 

390=\dfrac{2}{1-r}

 

1-r=\dfrac{2}{390}

 

r=1-\dfrac{2}{390}

 

r=\dfrac{388}{390}

 

r=\dfrac{194}{195}

The first option is an arithmetic series:

 

a=1

 

d=1

 

n=64

 

Substitute into formula:

 

\begin{aligned}S_{n}&=\dfrac{n}{2}(2a+(n-1)d)\\[1.2em]&=\dfrac{64}{2}(2\times 1+(64-1)1)\\[1.2em]&=32(2+63)\\[1.2em]&=32\times 65\\[1.2em]&=£2080\end{aligned}

 

The second option is a geometric series:

 

a=0.01

 

r=2

 

n=64

 

Substitute into formula:

 

\begin{aligned}S_{n}&=\dfrac{a(1-r^{n})}{1-r}\\[1.2em]&=\dfrac{0.01(1-2^{64})}{1-2}\\[1.2em]&=£1.84\times 10^{17}\end{aligned}

 

 

So the second chessboard is far more valuable.