SATHEE: Binomial-Theorem-And-Its-Simple-Applications Question 246

Binomial-Theorem-And-Its-Simple-Applications Question 246

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} $

Show Answer

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} $

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Binomial-Theorem-And-Its-Simple-Applications Question 246

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:

Answer:

Solution:

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