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  • <img alt="\small \frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+\ldots=" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Csmall%20%5Cfrac%7BC_0%7D%7B1%7D&plus;%5Cfrac%7BC_2%7D%7B3%7D&plus

\small \frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+\ldots=

Option: 1

\frac{2^n}{n}


Option: 2

2^n


Option: 3

\frac{2^n}{n+1}


Option: 4

None of these


Answers (1)

Consider the expansion
(1+x)^n=C_0+C_1 x+C_2 x^2+C_3 x^3+C_4 x^4+\ldots+C_n x^n

Integrating both sides of (i) within limits –1 to 1, we get

\begin{aligned} & \int_{-1}^1(1+x)^n d x=\int_{-1}^1\left(C_0+C_1 x+C_2 x^2+C_3 x^3+C_4 x^4+\ldots C_n x^n\right) d x \\ & =\int_{-1}^1\left(C_0+C_2 x^2+C_4 x^4+\ldots\right) d x \int_{-1}^1\left(C_1 x+C_3 x^3+\ldots .\right) d x \end{aligned}

=2 \int_{-1}^1\left(C_0+C_2 x^2+C_4 x^4+\ldots\right) d x+0

(By Prop. Of definite integral)(since second integral contains odd function)
\left.\left.\frac{(1+x)^{n+1}}{n+1}=\right]_{-1}^1=2\left(C_0 x+\frac{C_2 x^3}{3}+\frac{C_4 X^5}{5}+\ldots\right)\right]_0^1

\frac{2^{n+1}}{n+1}=2\left(C_0 \frac{C_2}{3}+\frac{C_4}{5}+\ldots\right)

Hence,  C_0 \frac{C_2}{3}+\frac{C_4}{5}+\ldots=\frac{2^n}{n+1}

Alternative Method.
\begin{aligned} \text { L.H.S. } & =C 0=\frac{C_2}{3}+\frac{C_4}{5}+\ldots \\ & =1+\frac{n(n-1)}{1.2 .3}+\frac{n(n-1)(n-2)(n-2)}{1.2 .3 .4 .5}+\ldots \\ & =\frac{1}{(n+1)}\left\{\frac{(n+1)}{1}+\frac{(n+1) n(n-1)}{1.2 .3}+\frac{(n+1) n(n-1)(n-2)(n-3)}{1.2 .3 .4 .5}+\ldots\right\} \\ & =\frac{1}{(n+1)}\left\{{ }^{n+1} C_1+{ }^{n+1} C_3+{ }^{n+1} C_5+\ldots\right\} \end{aligned}

=\frac{1}{(n+1)}           {sum of even binomial coefficients of (1 + x)n+1}

=\frac{2^{n+1-1}}{(n+1)}
=\frac{2^n}{n+1}     = R. H. S. coefficient divided by an integer i.e. in the   form of  \frac{{ }^n C_r}{k}.
 

Posted by

Kshitij

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