Question

The value of $\sum _{r=0}^{10}$ ${}^{20}{C}_r$ is equal to?

• A. $\frac{1}{2}(2^{20}+$ ${}^{20}{C}_{10})$
• B. $\frac{1}{2}(2^{28}+$ ${}^{19}{C}_{10})$
• C. $20(2^{18}+$ ${}^{19}{C}_{11})$
• D. $10(2^{18}+$ ${}^{19}{C}_{11})$

Answer 1

Answer: (A)

$\frac{1}{2}(2^{20}+$ ${}^{20}{C}_{10})$

$\begin{array}{c}{\sum }_{r=0}^{10}{}^{20}{C}_r={}^{20}{C}_0+{}^{20}{C}_1........{}^{20}{C}_r=2^{20}\\ \left({}^{20}{C}_0{+}^{20}{C}_1{....}^{20}{C}_{10}\right)+\left({}^{20}{C}_{11}{......}^{20}{C}_{20}\right)=2^{20}\\ \left({}^{20}{C}_0{......}^{20}{C}_{10}\right)+\left({}^{20}{C}_9{+}^{20}{C}_8{.....}^{20}{C}_0\right)=2^{20}\\ =2\left({}^{20}{C}_0{........}^{20}{C}_{10}\right)=2^{20}{+}^{20}{C}_{10}\\ {=}^{20}{C}_0+...{.}^{20}{C}_{10}=2^{19}+{\frac{1}{2}}^{20}{C}_{10}\end{array}$