Question

Sum of $n$ terms of the series$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+......$ is

• A. $2^{-\mathrm{n}}$
• B. $2^{-\mathrm{n}}(\mathrm{n}–1)$
• C. $2^{\mathrm{n}}(\mathrm{n}–1)+1$
• D. $2^{-\mathrm{n}}+\mathrm{n}–1$

Answer 1

The correct option is D

$2^{-\mathrm{n}}+\mathrm{n}–1$

Explanation for the correct option:

Finding the sum of ‘$n$’ terms of the series:

Given series are $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+......$

Let the sum of ‘$n$’ terms be ${\mathrm{S}}_{\mathrm{n}}$ .

Therefore,

$$\begin{gathered} {\mathrm{S}}_{\mathrm{n}}=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+......+n^{th}\text{term} \\ =\frac{\left(2-1\right)}{2}+\frac{\left(4-1\right)}{4}+\frac{\left(8-1\right)}{8}+\frac{\left(16-1\right)}{16}+.......... \\ =\left(\frac{2}{2}-\frac{1}{2}\right)+\left(\frac{4}{4}-\frac{1}{4}\right)+\left(\frac{8}{8}-\frac{1}{8}\right)+\left(\frac{16}{16}-\frac{1}{16}\right)+.......... \\ =1-\frac{1}{2}+1-\frac{1}{4}+1-\frac{1}{8}+1-\frac{1}{16}+......... \\ =\left(1+1+1+1+.........\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+........\right)\to \left(i\right) \end{gathered}$$

Now, $\frac{1}{2},\frac{3}{4},\frac{7}{8},\frac{15}{16}$are terms of G.P. with common ratio $\mathrm{r}=\frac{1}{2}$ and first term $a=\frac{1}{2}$

The sum of terms of G.P be $\frac{a\left(1-r^n\right)}{1-r}$ where $a$ is the first term and $r$ is the common ratio where $r<1$

. $$\begin{gathered} \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+........+n^{th}\text{term}=\frac{\frac{1}{2}\times \left(1-{\left(\frac{1}{2}\right)}^{\mathrm{n}}\right)}{\left(1-\frac{1}{2}\right)} \\ =\frac{\frac{1}{2}\times \left(1-{\left(\frac{1}{2}\right)}^{\mathrm{n}}\right)}{\frac{1}{2}} \\ =1-{\left(\frac{1}{2}\right)}^{\mathrm{n}} \end{gathered}$$

Then equation $\left(i\right)$ becomes

$$\begin{gathered} {\mathrm{S}}_{\mathrm{n}}=\left(1+1+1+........\right)-\left(1-{\left(\frac{1}{2}\right)}^{\mathrm{n}}\right) \\ =\mathrm{n}-1+{\left(\frac{1}{2}\right)}^{\mathrm{n}} \\ =\mathrm{n}-1+2^{-\mathrm{n}} \end{gathered}$$

Hence, option (D) is the correct answer.