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 $2^{19}+\frac{1}{2}{.}^{20}{C}_{10}$

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

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

and since ${{}^{20}C}_{20}={{}^{20}C}_0,{{}^{20}C}_{19}={{}^{20}C}_1,{{}^{20}C}_{18}={{}^{20}C}_2...{{}^{20}C}_{11}={{}^{20}C}_9$

Thus

$\sum _{r=0}^{10}{{}^{20}C}_r=({{}^{20}C}_0+{{}^{20}C}_1+...+{{}^{20}C}_{20})-({{}^{20}C}_{11}+{{}^{20}C}_{12}+...+{{}^{20}C}_{20})$

$=2^{20}-({{}^{20}C}_{11}+{{}^{20}C}_{12}+...+{{}^{20}C}_{20})$

$=2\left[({{}^{20}C}_0+{{}^{20}C}_1+...+{{}^{20}C}_{10})\right]=2^{20}+{{}^{20}C}_{10}$

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