Question

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

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

Answer 1

The correct option is B $2^{19}+\frac{1}{2}.{}^{20}{C}_{10}$

We have, ${\sum }_{r=0}^{10}{}^{20}{C}_r{=}^{20}{C}_0{+}^{20}{C}_1+...{+}^{20}{C}_{10}$

But ${}^{20}{C}_0{+}^{20}{C}_1+...{+}^{20}{C}_{20}=2^{20}$

and $\because {}^{20}{C}_{20}{=}^{20}{C}_0{;}^{20}{C}_{19}{=}^{20}{C}_1{;}^{20}{C}_{18}{=}^{20}{C}_2...$ and ${}^{20}{C}_{11}{=}^{20}{C}_9$

$\therefore \sum _{r=0}^{10}{}^{20}{C}_r=({}^{20}{C}_0{+}^{20}{C}_1+...{+}^{20}{C}_{20})-({}^{20}{C}_{11}{+}^{20}{C}_{12}+...{+}^{20}{C}_{20})=2^{20}{+}^{20}{C}_{10}-({}^{20}{C}_{10}{+}^{20}{C}_9+...{+}^{20}{C}_0)\Rightarrow 2[{}^{20}{C}_0{+}^{20}{C}_1+...{+}^{20}{C}_{10}]=2^{20}{+}^{20}{C}_{10}$

${\therefore }^{20}{C}_0{+}^{20}{C}_1+...{+}^{20}{C}_{10}=2^{19}+\frac{1}{2}{.}^{20}{C}_{10}$