Question

$\sum _{r=0}^n(-1{)}^r{}^n{C}_r[\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+\cdots \infty ]$ is equal to

• A. $\frac{1}{2^n-1}$
• B. $\frac{1}{1-2^n}$
• C. $\frac{1}{2^n+1}$
• D. $\frac{2^n}{2^n+1}$

Answer 1

The correct option is A $\frac{1}{2^n-1}$

$\sum _{r=0}^n(-1{)}^r{}^n{C}_r{\left(\frac{1}{2}\right)}^r={(1-\frac{1}{2})}^n={\left(\frac{1}{2}\right)}^n$

$\sum _{r=0}^n(-1{)}^r{}^n{C}_r{\left(\frac{3}{4}\right)}^r={(1-\frac{3}{4})}^n={\left(\frac{1}{4}\right)}^n$

$\sum _{r=0}^n(-1{)}^r{}^n{C}_r{\left(\frac{7}{8}\right)}^r={(1-\frac{7}{8})}^n={\left(\frac{1}{8}\right)}^n$

$⋮$
and so on.

$\therefore \sum _{r=0}^n(-1{)}^r{}^n{C}_r[\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+\cdots \infty ]$

$={\left(\frac{1}{2}\right)}^n+{\left(\frac{1}{4}\right)}^n+{\left(\frac{1}{8}\right)}^n+\cdots \infty$

$=\frac{1}{2^n}+\frac{1}{2^{2n}}+\frac{1}{2^{3n}}+\cdots \infty$
Above series is a G.P. with $a=r=\frac{1}{2^n}$
So, sum $=\frac{\frac{1}{2^n}}{1-\frac{1}{2^n}}[\because S_{\infty }=\frac{a}{1-r}]$

$=\frac{1}{2^n-1}$