If $\frac{1}{2 imes 4}+\frac{1}{4 imes 6}+\frac{1}{6 imes 8}+\ldots(n$ terms $)=\frac{k n}{n imes 1}$, then $k$ is equal to

Question

If $\frac{1}{2 	imes 4}+\frac{1}{4 	imes 6}+\frac{1}{6 	imes 8}+\ldots(n$ terms $)=\frac{k n}{n 	imes 1}$, then $k$ is equal to

Aicharya · Accepted Answer

Given series can be rewritten as $\begin{gathered}\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\ldots+\frac{1}{2 n}-\frac{1}{2 n+2}ight] \ \quad=\frac{1}{4}\left(1-\frac{1}{n+1}ight) \ =\frac{1}{4}\left(\frac{n}{n+1}ight)\end{gathered}$ On comparing given equation, we get $k=\frac{1}{4}$

If $\frac{1}{2 \times 4}+\frac{1}{4 \times 6}+\frac{1}{6 \times 8}+\ldots(n$ terms $)=\frac{k n}{n \times…

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