Question

The sum of the series $\sum _{r=0}^{10}{}^{20}{C}_r$ is

• A. $2^{19}-\frac{1}{2}\cdot {}^{20}{C}_{10}$
• B. $2^{19}+\frac{1}{2}\cdot {}^{20}{C}_{10}$
• C. $2^{19}$
• D. $2^{20}$

Answer 1

Answer: (B)

We have, $\sum _{r=0}^{10}{}^{20}{C}_r={}^{20}{C}_0+{}^{20}{C}_1+\dots +{}^{20}{C}_{10}$
But ${}^{20}{C}_0+{}^{20}{C}_1+\dots {}^{20}{C}_{20}=2^{20}$
and $\because {}^{20}{C}_{20}={}^{20}{C}_0,{}^{20}{C}_{19}={}^{20}{C}_1$
${}^{20}{C}_{18}={}^{20}{C}_2,{}^{20}{C}_{11}={}^{20}{C}_9$
$\therefore \sum _{r=0}^{10}{}^{20}{C}_r=\left({}^{20}{C}_0+{}^{20}{C}_1+\dots +{}^{20}{C}_{20}\right)$
$-\left({}^{20}{C}_{11}+{}^{20}{C}_{12}+\dots +{}^{20}{C}_{20}\right)$
$=2^{20}+{}^{20}{C}_{10}-\left({}^{20}{C}_{10}+{}^{20}{C}_9+\dots +{}^{20}{C}_0\right)$
$\Rightarrow 2\left({}^{20}{C}_0+{}^{20}{C}_1+\dots +{}^{20}{C}_{10}\right)=2^{20}+{}^{20}{C}_{10}$
$\therefore {}^{20}{C}_0+{}^{20}{C}_1+\dots +{}^{20}{C}_{10}=2^{19}+\frac{1}{2}{}^{20}{C}_{10}$