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  • <img alt="\frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+\frac{C_6}{7}+\ldots=" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7BC_0%7D%7B1%7D&plus;%5Cfrac%7BC_2%7D%7B3%7D&plus;%5C

\frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+\frac{C_6}{7}+\ldots=

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

None of these


Answers (1)

best_answer

Putting the values of C_0, C_2, C_4 \ldots we get

=1+\frac{n(n-1)}{3.2 !}+\frac{n(n-1)(n-2)(n-3)}{5.4 !}+\ldots

=\frac{1}{n+1}\left[(n+1)+\frac{(n+1) n(n-1)}{3 !}+\frac{(n+1) n(n-1)(n-2)(n-3)}{5 !}+\ldots\right]....(i)

Put n+1 =N

=\frac{1}{N}\left[N+\frac{N(N-1)(N-2)}{3 !}+\frac{N(N-1)(N-2)(N-3)(N-4)}{5 !}+\ldots\right]..(ii)

=\frac{1}{N}\left\{{ }^N C_1+{ }^N C_3+{ }^N C_5+\ldots .\right\}..(iii)

=\frac{1}{N}\left\{2^{N-1}\right\}=\frac{2^n}{n+1} \quad\{N=n+1\}

Trick : Put n=1, then S_1=\frac{{ }^1 C_0}{1}=\frac{1}{1}=1

\begin{aligned} & \text { At } n=2, \quad S_2=\frac{{ }^2 C_0}{1}+\frac{{ }^2 C_2}{3}=1+\frac{1}{3}=\frac{4}{3} \\ & \Rightarrow S_1=1, S_2=\frac{4}{3} \end{aligned}

Option (c) satisfies these values

 

 

Posted by

Pankaj

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