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  • If  <img alt="C_r" src="https://entrancecorner.oncodecogs.com/gif.latex?

If  C_r stands for { }^{\mathrm{n}} C_r \text {, } the sum of given series \frac{2(n / 2) !(n / 2) !}{n !}\left[C_0^2-2 C_1^2+3 C_2^2-\ldots \ldots+(-1)^n(n+1) C_n^2\right]

where n is an even positive integer, is

Option: 1

0


Option: 2

(-1)^{n / 2}(n+1)


Option: 3

(-1)^n(n+2)


Option: 4

(-1)^{n / 2}(n+2)


Answers (1)

best_answer

We have C_0^2-2 C_1^2+3 C_2^2-\ldots \ldots \ldots+(-1)^n(n+1) C_n^2=\left[C_0^2-C_1^2+C_2^2-\ldots \ldots .+(-1)^n C_n^2\right]-\left[C_1^2-2 C_2^2+3 C_3^2 \ldots \ldots+(-1)^n n \cdot C_n^2\right]

$$ \begin{aligned} & =\frac{2 \cdot \frac{n}{2} ! \frac{n}{2} !}{n !}\left[(-1)^{n / 2} \cdot\left(1+\frac{n}{2}\right) \frac{n !}{\frac{n}{2} ! \frac{n}{2} !}\right]=(-1)^{n / 2}(n+2) \\ & \end{aligned}Therefore the value of given expression

Posted by

Ritika Harsh

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