Question

Find the sum of the series $\displaystyle \sum_{r=0}^n (-1)^r \,{}^nC_r \, \bigg[ \frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+\frac{15^r}{2^{4r}}......... $ upto $m $ terms $\bigg]$

Answer 1

$\displaystyle \sum_{r=0}^n(-1)^r \,{}^nC_r \, \bigg[ \frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+\frac{15^r}{2^{4r}}.........$ upto $m $ terms $\bigg]$
$ = \displaystyle \sum_{r=0}^n (-1)^r \,{}^nC_r \, \bigg(\frac{1}{2}\bigg)^r+ \displaystyle \sum_{r=0}^n (-1)^r \,{}^nC_r \, \bigg(\frac{3}{4}\bigg)^r$
$\displaystyle \sum_{r=0}^n (-1)^r \,{}^nC_r \, \bigg(\frac{7}{8}\bigg)^r+.....$ upto $ m $ terms
$=\bigg(1-\frac{1}{2}\bigg)^n+\bigg(1-\frac{3}{4}\bigg)^n+\bigg( 1-\frac{7}{8}\bigg)^n+... $ upto $m $ terms
$\bigg[ $ using $ \displaystyle \sum_{r=0}^n (-1)^r {}^{n}C_r x^{r}=(1+x)^n\bigg]$
$=\bigg(\frac{1}{2}\bigg)^n+\bigg(\frac{1}{4}\bigg)^n+\bigg(\frac{1}{8}\bigg)^n+......$ upto $m$ terms
$=\bigg(\frac{1}{2}\bigg)^n \Bigg[\frac{1-\bigg(\frac{1}{2^n}\bigg)^m}{1-\frac{1}{2^n}}\Bigg]=\frac{2^{mn}-1}{2^{mn}(2^n-1)}$