$1. {C_{0}}^{2} + 3. {C_{1}}^{2} + 5. {C_{2}}^{2} + . . . + (2 n + 1) . {C_{n}}^{2} = 2 n .^{2 n - 1} C_{n} +^{2 n} C_{n} .$

True

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False

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Solution

The correct option is A True

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Similar questions

C_{0}^{2} + 3. C_{1}^{2} + 5. C_{2}^{2} + \dots \dots + (2 n + 1) . C_{n}^{2} =

C_{0}^{2} + \frac{C_{1}^{2}}{2} + \frac{C_{2}^{2}}{3} + \dots \dots + \frac{C_{n}^{2}}{n + 1} = \frac{(2 n + 1)!}{[(n + 1)!)^{2}}

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

^{2 n} C_{n} =

x

(1 + x)^{n} {(1 + \frac{1}{x})}^{n}

^{2 n} C_{n} = \frac{1.3.5.7 \dots \dots (2 n - 1)}{n!}

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

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 - -}

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