Question

Evalute
(v) ${\sum }_{r=1}^5$ ${}^5{C}_r$

Answer 1

We have,

${\sum }_{r=1}^5$ ${}^5{C}_r={}^5{C}_1+{}^5{C}_2+{}^5{C}_3+{}^5{C}_4+{}^5{C}_5$

Using ${}^n{C}_r=\frac{n!}{r!(n-r)!}$

$\Rightarrow {\sum }_{r=1}^5$${}^5{C}_r=5+10+10+5+1$

$\Rightarrow {\sum }_{r=1}^5$${}^5{C}_r=31$

Hence, the sum of the series is 31.