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Q. The sum of the series $\displaystyle\sum_{r=0}^{10}$$\,{}^{20}C_r \,is$

Binomial Theorem

Solution:

$\sum\limits^{10}_{r = 0} \,{}^{20}C_{r} = \,{}^{20}C_{0} + \,{}^{20}C_{1} + ........... + \,{}^{20}C_{10}$
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 $^{20}C_{11} = \,{}^{20}C_{9}$
$\therefore \,{}^{20}C_{0}+\,{}^{20}C_{1} + ..... + \,{}^{20}C_{10}$
$= \left(^{20}C_{11}+\,{}^{20}C_{12}+ ..... +\,{}^{20}C_{20}\right)$
$-\left(^{20}C_{11}+\,{}^{20}C_{12}+ ..... +\,{}^{20}C_{20}\right)$
$= 2^{20}+\,{}^{20}C_{10}-\left(^{20}C_{10}+\,{}^{20}C_{9}+ ..... +\,{}^{20}C_{0}\right)$
$\therefore 2\left[^{20}C_{0}+\,{}^{20}C_{1}+..... + \,{}^{20}C_{10}\right]= 2^{20}+\,{}^{20}C_{10}$
$\therefore \,{}^{20}C_{0}+\,{}^{20}C_{1}+..... +\,{}^{20}C_{10} = 2^{19}+\frac{1}{2} \,{}^{20}C_{10}$