Question

Find the sum of the series $\sum _{r=0}^n(-1{)}^r{}^n{C}_r[\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+\frac{15^r}{2^{4r}}.........$ upto $m$ terms $]$

Answer 1

$\sum _{r=0}^n(-1{)}^r{}^n{C}_r[\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+\frac{15^r}{2^{4r}}.........$ upto $m$ terms $]$
$=\sum _{r=0}^n(-1{)}^r{}^n{C}_r(\frac{1}{2}{)}^r+\sum _{r=0}^n(-1{)}^r{}^n{C}_r(\frac{3}{4}{)}^r$
$\sum _{r=0}^n(-1{)}^r{}^n{C}_r(\frac{7}{8}{)}^r+.....$ upto $m$ terms
$=(1-\frac{1}{2}{)}^n+(1-\frac{3}{4}{)}^n+(1-\frac{7}{8}{)}^n+...$ upto $m$ terms
$[$ using $\sum _{r=0}^n(-1{)}^r{}^n{C}_r{x}^r=(1+x{)}^n]$
$=(\frac{1}{2}{)}^n+(\frac{1}{4}{)}^n+(\frac{1}{8}{)}^n+......$ upto $m$ terms
$=\left(\frac{1}{2}{)}^n\right[\frac{1-(\frac{1}{2^n}{)}^m}{1-\frac{1}{2^n}}]=\frac{2^{mn}-1}{2^{mn}(2^n-1)}$