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7.Binomial Theorem
hard
$\sum_{\mathrm{k}=0}^{20}\left({ }^{20} \mathrm{C}_{\mathrm{k}}\right)^{2}$ is equal to :
A
${ }^{40} \mathrm{C}_{21}$
B
${ }^{40} \mathrm{C}_{19}$
C
${ }^{40} \mathrm{C}_{20}$
D
${ }^{41} \mathrm{C}_{20}$
(JEE MAIN-2021)
Solution
$\sum_{\mathrm{k}=0}^{20}{ }^{20} \mathrm{C}_{\mathrm{k}} \cdot{ }^{20} \mathrm{C}_{20-\mathrm{k}}$
sum of suffix is const. so summation will be ${ }^{40} \mathrm{C}_{20}$
Standard 11
Mathematics