Question

How do you find the sum of finite geometric series?

Answer 1

Finding the sum of a finite geometric series:

Step-1: Assumption:

Let us consider a finite geometric series of $\left(n+1\right)$ terms whose first term is $a$ and the common ratio is $r$: $a,ar,a{r}^2,a{r}^3,\dots ,a{r}^n$.

Suppose $S_n$ be the sum of this series.

Step-2: Calculating $S_n$

We have, $S_n=a+ar+a{r}^2+\cdots +a{r}^n$.

Then,

$$\begin{gathered} r{S}_n=r\left(a+ar+a{r}^2+\cdots a{r}^n\right) \\ =ar+a{r}^2+a{r}^3+\cdots +a{r}^{n+1} \end{gathered}$$

Thus, we get:

$$\begin{gathered} r{S}_n-S_n=\left(ar+a{r}^2+a{r}^3+\cdots +a{r}^{n+1}\right)-\left(a+ar+a{r}^2+\cdots +a{r}^n\right) \\ =ar+a{r}^2+a{r}^3+\cdots +a{r}^{n+1}-a-ar-a{r}^2-\cdots -a{r}^n \\ =a{r}^{n+1}-a \\ \Rightarrow S_n\left(r-1\right)=a\left(r^{n+1}-1\right) \\ \Rightarrow S_n=\frac{a\left(r^{n+1}-1\right)}{r-1} \end{gathered}$$

Therefore, the sum of a finite geometric series of $\left(n+1\right)$ terms whose first term is $a$ and the common ratio is $r$ is given by: $S_n=\frac{a\left(r^{n+1}-1\right)}{r-1}$.