The value of the sum ${({^{n} C}_{1})}^{2} + {({^{n} C}_{2})}^{2} + {({^{n} C}_{3})}^{2} + . . . . . . {({^{n} C}_{n})}^{2}$ is

{({^{2 n} C}_{n})}^{2}

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({^{2 n} C}_{n})

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({^{2 n} C}_{n}) + 1

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({^{2 n} C}_{n}) - 1

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Solution

The correct option is A $({^{2 n} C}_{n}) - 1$
${(1 + x)}^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + . . . +^{n} C_{n} x^{n} . . . (i) {(x + 1)}^{n} =^{n} C_{0} x^{n} +^{n} C_{1} x^{n - 1} + . . . +^{n} C_{n} . . . (i i)$
Multiplying $(i)$ and $(i i)$ ,
${(1 + x)}^{2 n} = {(^{n} C_{0})}^{2} + {(^{n} C_{1})}^{2} + {(^{n} C_{2})}^{2} + . . . + {(^{n} C_{n})}^{2}$
Coefficient of $x^{n}$ in $(1 + x)^{2 n}$ will be $^{2 n} C_{n} - 1$

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