**Question:**

The sum of $n$ terms of a progression is $\left(2^{n}-1\right)$. Show that it is a GP and find its common ratio

**Solution:**

In this question, we will try to rewrite the given sum of the progression like the formula for the sum a G.P. series.

It is given that $S_{n}=\left(2^{n}-1\right)$

The formula for the sum of a G.P. series is,

$\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}-1}$

By solving the 2 equations together, we can say that

$\left(2^{n}-1\right)=a \frac{r^{n}-1}{r-1}$

$\Rightarrow 1 \times \frac{\left(2^{\mathrm{n}}-1\right)}{2-1}=a \frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}-1}$

By corresponding the numbers with the variables, we can conclude

$a=1$

r = 2

The G.P. series will therefore look like ⇒ 1,2,4,8,16,……to n terms

∴ The given progression is a G.P. series with the common ration being 2.