Question:                         If   $ \frac{1}{2\times 4}+\frac{1}{4\times 6}+\frac{1}{6\times 8}+…n $    terms   $ =\frac{kn}{n+1}, $    then k is equal to
Options:
A)   $ \frac{1}{4} $
B)   $ \frac{1}{2} $
C) 1
D)   $ \frac{1}{8} $
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  Answer:
Correct Answer: A
Solution:
- [a]   $ \frac{kn}{n+1}=[ \frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+…n,terms ] $      $ =\frac{1}{2}[ \frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+…+\frac{2n+2-2n}{2n(2n+2)} ] $      $ =\frac{1}{2}[ \frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+…+\frac{1}{2n}-\frac{1}{2n+2} ] $      $ =\frac{1}{2}[ \frac{1}{2}-\frac{1}{2(n+1)} ]=\frac{n}{4(n+1)}\Rightarrow k=\frac{1}{4} $