Question

The sum of the series $\displaystyle\sum_{r=0}^{10}{ }^{20} C_{r}$ is

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

Answer 1

Answer: (B)

We have, $\displaystyle\sum_{r=0}^{10}{ }^{20} C_{r}={ }^{20} C_{0}+{ }^{20} C_{1}+\ldots+{ }^{20} C_{10}$
But ${ }^{20} C_{0}+{ }^{20} C_{1}+\ldots{ }^{20} C_{20}=2^{20}$
and $\because { }^{20} C_{20} ={ }^{20} C_{0},{ }^{20} C_{19}={ }^{20} C_{1} $
${ }^{20} C_{18} ={ }^{20} C_{2},{ }^{20} C_{11}={ }^{20} C_{9}$
$\therefore \displaystyle\sum_{r=0}^{10}{ }^{20} C_{r}=\left({ }^{20} C_{0}+{ }^{20} C_{1}+\ldots+{ }^{20} C_{20}\right)$
$-\left({ }^{20} C_{11}+{ }^{20} C_{12}+\ldots+{ }^{20} C_{20}\right)$
$=2^{20}+{ }^{20} C_{10}-\left({ }^{20} C_{10}+{ }^{20} C_{9}+\ldots+{ }^{20} C_{0}\right)$
$\Rightarrow 2\left({ }^{20} C_{0}+{ }^{20} C_{1}+\ldots+{ }^{20} C_{10}\right)=2^{20}+{ }^{20} C_{10}$
$\therefore { }^{20} C_{0}+{ }^{20} C_{1}+\ldots+{ }^{20} C_{10}= 2 ^{19}+\frac{ 1 }{ 2 }{ }^{20} C_{10}$