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If $\frac{1}{2 \times 4}+\frac{1}{4 \times 6}+\frac{1}{6 \times 8}+\ldots(n$ terms $)=\frac{k n}{n \times 1}$, then $k$ is equal to
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1647 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{4}$
Given series can be rewritten as
$\begin{gathered}\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\ldots+\frac{1}{2 n}-\frac{1}{2 n+2}\right] \\ \quad=\frac{1}{4}\left(1-\frac{1}{n+1}\right) \\ =\frac{1}{4}\left(\frac{n}{n+1}\right)\end{gathered}$
On comparing given equation, we get
$k=\frac{1}{4}$
$\begin{gathered}\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\ldots+\frac{1}{2 n}-\frac{1}{2 n+2}\right] \\ \quad=\frac{1}{4}\left(1-\frac{1}{n+1}\right) \\ =\frac{1}{4}\left(\frac{n}{n+1}\right)\end{gathered}$
On comparing given equation, we get
$k=\frac{1}{4}$
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