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  4. The sum of the series 20C0-20C1+20C2-20C3+ ldots +20C10 is equal to

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Q. The sum of the series ${}^{20}C_{0}-{}^{20}C_{1}+{}^{20}C_{2}-{}^{20}C_{3}+\ldots +{}^{20}C_{10}$ is equal to

NTA AbhyasNTA Abhyas 2022

Solution:

$S={ }^{20} C_{0}-{ }^{20} C_{1}+{ }^{20} C_{2}+\ldots+{ }^{20} C_{10}$
We know that,
${ }^{20} C_{0}-{ }^{20} C_{1}+{ }^{20} C_{2}-\ldots+{ }^{20} C_{20}=0 $
$\Rightarrow 2^{20} C_{0}-{ }^{20} C_{1}+{ }^{20} C_{2}+\ldots-{ }^{20} C_{9}+{ }^{20} C_{10}=0$
$\Rightarrow{ }^{20} C_{0}-{ }^{20} C_{1}+{ }^{20} C_{2} \ldots-{ }^{20} C_{9}=-\frac{1}{2}{ }^{20} C_{10}$
So, $ S=-\frac{1}{2}{ }^{20} C_{10}+{ }^{20} C_{10}=\frac{1}{2}{ }^{20} C_{10}$