Sequence And Series Question 133
Question: Sum of first $ n $ terms in the following series $ {{\cot }^{-1}}3+{{\cot }^{-1}}7+{{\cot }^{-1}}13+{{\cot }^{-1}}21+…………. $ is given by
Options:
A) $ {{\tan }^{-1}}( \frac{n}{n+2} ) $
B) $ {{\cot }^{-1}}( \frac{n+2}{n} ) $
C) $ {{\tan }^{-1}}(n+1)-{{\tan }^{-1}}1 $
D) All of these
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ S=3+7+13+21+……..+T_{n} $
$ \Rightarrow $ $ T_{n}=n^{2}+n+1 $ . Let $ T_{r}={{\cot }^{-1}}(r^{2}+r+1)={{\tan }^{-1}}(r+1)-{{\tan }^{-1}}r $ . Put $ r=1,\ 2,………,n $ and add, we get the required sum $ {{\tan }^{-1}}(n+1)-{{\tan }^{-1}}1={{\tan }^{-1}}( \frac{n}{n+2} )={{\cot }^{-1}}( \frac{n+2}{n} ) $ .
BETA
