If ${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} . . . . . . . C_{n} x^{n}$ , then $C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + . . . . . . . + C_{n}^{2}$ is equal to

\frac{n!}{n! n!}

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\frac{(2 n)!}{n! n!}

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\frac{(2 n)!}{n!}

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Solution

The correct option is A $\frac{n!}{n! n!}$
$\begin{matrix} {(1 + x)}^{h} = C_{0} + C_{1} x + C_{2} x^{2} . . . . . . . . . e_{n} x^{4} {(1 + x)}^{h} = C_{0} x^{h} + C_{1} x^{h - 1} (1) \times (2) {(1 + x)}^{2 h} = (C_{0} + C_{1} x . . . . . C_{2} x^{4}) c o m p a r i s o n 2 h C_{n} = C_{0}^{2} + C_{1}^{2} . . . . . . C_{n}^{2} C_{0}^{2} + C_{1}^{2} . . . . . . C_{n}^{2} = \frac{2 n!}{(n! n!)} \end{matrix}$

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Similar questions

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . . + C_{n} x^{2}$ , then

$C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + . . . . . . . . . . + C_{n}^{2}$ =

c_{0}, c_{1}, c_{2}, . . . . . . . c_{n}

{(1 + x)}^{n}

{c_{0}}^{2} + {c_{1}}^{2} + {c_{2}}^{2} + . . . . {c_{n}}^{2} = \frac{| 2 n - - -}{| n - - | n - -}

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . . + C_{n} x^{2}$ , then

$C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + . . . . . . . . . . + C_{n}^{2}$ =

{(1 + x)}^{n} = n \sum r = 0 C_{r} x^{r} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}

C_{0}^{2} - C_{1}^{2} + C_{2}^{2} - C_{3}^{2} + . . . (- 1)^{n} C_{n}^{2}

n = 15

{(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}

2 C_{0} + 2^{2} . \frac{C_{1}}{2} + 2^{3} . \frac{C_{2}}{3} + . . . + 2^{n + 1} . \frac{C_{n}}{n + 1} =

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