Question

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

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

Answer 1

Answer: (C)

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