Question

If $\frac{1}{2\times 4}+\frac{1}{4\times 6}+\frac{1}{6\times 8}+\dots n$ terms $=\frac{kn}{n+1}$, then $k$ is equal to

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. 1
• D. 8

Answer 1

Answer: (A)

$\frac{kn}{n+1}=[\frac{1}{2\cdot 4}+\frac{1}{4\cdot 6}+\frac{1}{6\cdot 8}+\dots n$ terms $]$
$=\frac{1}{2}\left[\frac{4-2}{2\cdot 4}+\frac{6-4}{4\cdot 6}+\frac{8-6}{6\cdot 8}+\dots +\frac{2n+2-2n}{2n(2n+2)}\right]$
$=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\dots +\frac{1}{2n}-\frac{1}{2n+2}\right]$
$=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{2(n+1)}\right]=\frac{n}{4(n+1)}$
$\Rightarrow k=\frac{1}{4}$