Question

Geometric Series assistance

Answer 1

we have been asked to find the sum of the series

$\sum _{n=1}^5\left(\frac{1}{3}\right)^{n-1}$

As we know that a geometric series has a constant ratio "r" and it is defined as

$r=\frac{a_{n+1}}{a_n}=\frac{\left(\frac{1}{3}\right)^{\left(n+1\right)-1}}{\left(\frac{1}{3}\right)^{n-1}}=\frac{1}{3}$

The first term of the series is $a_1=\left(\frac{1}{3}\right)^{1-1}=1$

Geometric series sum formula is

$S_n=a_1\frac{1-r^n}{1-r}$

Plugin the values we get

$S_5=1\cdot \frac{1-\left(\frac{1}{3}\right)^5}{1-\frac{1}{3}}$

On simplification we get

$S_5=\frac{121}{81}$

Hence the sum of the given series is $\frac{121}{81}$