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  • If   stands for <img alt="\small { }^n C_r" src="https://entrancecorner.oncodecogs.com/gif.lat

If  \small C_r stands for \small { }^n C_r, then the sum of the series

\small \frac{2\left(\frac{n}{2}\right) !\left(\frac{n}{2}\right) !}{n !}\left[C_0^2-2 C_1^2+\ldots \ldots \ldots+(-1)^n(n+1) C_n^2\right]

where n is an even positive integer, is equal to

Option: 1

0


Option: 2

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


Option: 3

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


Option: 4

None of these


Answers (1)

best_answer

When n is an even positive integer n is of the form
2K, K∈N 

\\Let\ f(K)=\frac{2 \underline{\underline{K}} \underline{\underline{K}}}{\lfloor\underline{K}}\left[C_0^2-2 C_1^2+3 C_2^2+\ldots\right. \left.+(-1)^{2 K}(2 K+1) C_{2 K}^2\right]

Reversing the order of terms

\begin{aligned} & f(K)=\frac{2 \underline{K} \mid K}{\underline{2 K}}\left[(2 K+1) C_0^2-2 K C_1^2+\ldots \ldots+C_{2 K}^2\right] \\ & \Rightarrow 2 f(K)=\frac{2 \underline{K} \mid \underline{K}}{\lfloor 2 K}(2 K+2)\left(C_0{ }^2-C_1{ }^2+C_2^2 \cdots \ldots . C_{2 K}^2\right) \\ & \left(\because C_0^2-C_1^2+C_2^2-\ldots+C_2 K^2=(-1)^{K 2 K} C_K\right) \\ & \Rightarrow 2 f(K)=\frac{2}{2 K_{C_K}} 2(K+1)(-1)^K 2{ }^2 C_K \\ & \therefore f(K)=(-1)^{n / 2}\left(\frac{n}{2}+1\right) 2=(-1)^{n / 2}(n+2) \text {. } \\ & \end{aligned}

Posted by

manish painkra

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