k = 0 20 ( C k 20 ) 2 is equal to

Question

$\sum_{k = 0}^{20} {({}^{20}C_{k})}^{2}$ is equal to

Smart Achievers · Accepted Answer

(3)

We know that

${(1 + x)}^{n} = {}^{n}C_{0} + {}^{n}C_{1} x + {}^{n}C_{2} x^{2} + \dots + {}^{n}C_{n} x^{n}$ ...(i)

Also, ${(x + 1)}^{n} = {}^{n}C_{0} x^{n} + {}^{n}C_{1} x^{n - 1} + {}^{n}C_{2} x^{n - 2} + \dots + {}^{n}C_{n}$ ...(ii)

Multiply (i) & (ii), we get

${(1 + x)}^{2 n} = ({}^{n}C_{0} + {}^{n}C_{1} x + \dots + {}^{n}C_{n} x^{n}) ({}^{n}C_{0} x^{n} + {}^{n}C_{1} x^{n - 1} + \dots + {}^{n}C_{n})$

Comparing coefficient of $x^{n}$ on both sides, we get

${}^{2 n}C_{n} = {({}^{n}C_{0})}^{2} + {({}^{n}C_{1})}^{2} + {({}^{n}C_{2})}^{2} + \dots + {({}^{n}C_{n})}^{2}$

$\Rightarrow \sum_{r = 0}^{n} {({}^{n}C_{r})}^{2} = {}^{2 n}C_{n}$

$∴$ $\sum_{k = 0}^{20} {({}^{20}C_{k})}^{2} = {}^{40}C_{20}$ $[∵ n = 20]$

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