Each term of the sum
$C_{0}^{2} + 3. C_{1}^{2} + 5. C_{2}^{2} . . . . . (2 n + 1) c_{n}^{2}$ can be written as $(2 r + 1)^{n} C_{r}^{2}$

True

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False

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Solution

The correct option is A
True

When we put r= 0,1,2,3...... we get $^{n} C_{0}^{2}, 3. C_{1}^{2} + 5. C_{2}^{2} . . . . . . .$ which are the terms of the sum.
$\Rightarrow$ Statement is true.

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Similar questions

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

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

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

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} .

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}}

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