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  • Find the sum of series <img alt="\frac{2\left(\frac{n}{2}\right) ! \cdot\left(\frac{n}{2}\right) !}{(n) !}\left(C_0^2-2 C_1^2+3 C_2^2-\ldots \ldots . .(-1)^

Find the sum of series \frac{2\left(\frac{n}{2}\right) ! \cdot\left(\frac{n}{2}\right) !}{(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.

Option: 1

(-1)^n(n+2)


Option: 2

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


Option: 3

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


Option: 4

(-1)^n(n+1)


Answers (1)

best_answer

(1+x)^n=C_0+C_1 x+\ldots \ldots \ldots+C_n x^n

Multiplying by x and differentiating with respect to x,

(1+x)^n+n x(1+x)^{n-1}=C_0+2 C_1 x+3 C_2 x^2+\ldots \ldots \ldots+(n+1) C_2 x^n

\frac{1}{x} \Rightarrow\left(1-\frac{1}{x}\right)^n=C_0-\frac{C_1}{x}+\frac{C_2}{x^2}-\ldots . .+\frac{(-1)^n C_n}{x^n}From (2) and (3), coefficient of term independent of x

\text { In }\left\{\frac{(1-x)^n}{x^n}\left[(1+x)^n+n x(1+x)^{n-1}\right]\right\}=\left[C_0^2-2 C_1^2+3 C_2^2-4 C_3^2 \ldots \ldots+(-1)^n(n+1) C_n^2\right] \\

=C_0^2-2 C_1^2+3 C_2^2-4 C_3^2+\ldots \ldots \ldots(-1)^n(n+1) C_n^2=coefficient of  x^n in 

{\left[\left(1-x^2\right)^n+n x\left(1-x^2\right)^{n-1}(1-x)\right]} \\

=(-1)^{n / 2}\,{ }^n C_{n / 2}-n{ }^{n-1} C_{(n-2) / 2}(-1)^{(n-2) / 2} \\
\Rightarrow \frac{n !}{(n / 2) ! \cdot(n / 2) !} \cdot\left(\frac{n+2}{2}\right) \\

\Rightarrow \frac{n(n / 2) ! \cdot(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)

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

 

Posted by

vishal kumar

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