Question

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

Answer 1

$$\begin{gathered} \mathrm{for}\sum _{r=0}^{10}{}^{20}{C}_r \\ \mathrm{for}n=2 \\ \mathrm{consider}, \\ {(1+x)}^{20}={}^{20}{C}_0+{}^{20}{C}_{1}x+......+{}^{20}{C}_{20}{x}^{20} \\ \mathrm{i}.\mathrm{e}.\sum _{r=0}^{10}{}^{20}{C}_r={}^{20}{C}_0+{}^{20}{C}_1+{}^{20}{C}_2+......+{}^{20}{C}_{10} \\ \mathrm{Since}{}^{20}{C}_1={}^{20}{C}_{19} \\ {}^{20}{C}_2={}^{20}{C}_{18} \\ . \\ . \\ . \\ {}^{20}{C}_{11}={}^{20}{C}_9 \\ \mathrm{i}.\mathrm{e}.{(1+x)}^{20}=2({}^{20}{C}_0+{}^{20}{C}_{1}x+.......+{}^{20}{C}_9{x}^9+{}^{20}{C}_{10}{x}^{10})-{}^{20}{C}_{10}{x}^{10} \\ {(1+x)}^{20}=2\left(\sum _{r=0}^{10}{}^{20}{C}_r{x}^r\right){-}^{20}{C}_{10}{x}^{10} \\ 2\left(\sum _{r=0}^{10}{}^{20}{C}_r{x}^r\right) \\ \mathrm{put}x=1 \\ \mathrm{i}.\mathrm{e}.2\sum _{r=0}^{10}{}^{20}{C}_r={(1+1)}^{20}{+}^{20}{C}_{10} \\ \mathrm{i}.\mathrm{e}.\sum _{r=0}^{10}{}^{20}{C}_r=2^{19}+\frac{{}^{20}{C}_{10}}{2} \end{gathered}$$