Sequence And Series Question 528
Question: The sum of the series $ \frac{1}{1+1^{2}+1^{4}}+\frac{2}{1+2^{2}+2^{4}}+\frac{3}{1+3^{2}+3^{4}}+……… $ to $ n $ terms is
Options:
A) $ \frac{n(n^{2}+1)}{n^{2}+n+1} $
B) $ \frac{n(n+1)}{2(n^{2}+n+1)} $
C) $ \frac{n(n^{2}-1)}{2(n^{2}+n+1)} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ T_{n} $ be the $ n^{th} $ term of the series $ \frac{1}{1+1^{2}+1^{4}}+\frac{2}{1+2^{2}+2^{4}}+\frac{3}{1+3^{2}+3^{4}}+…… $ Then $ T_{n}=\frac{n}{1+n^{2}+n^{4}}=\frac{n}{{{(1+n^{2})}^{2}}-n^{2}} $ $ =\frac{n}{(n^{2}+n+1)(n^{2}-n+1)} $ $ =\frac{1}{2}[ \frac{1}{n^{2}-n+1}-\frac{1}{n^{2}+n+1} ] $ $ =\frac{1}{2}[ \frac{1}{1+(n-1)n}-\frac{1}{1+n(n+1)} ] $ Now $ \sum\limits_{r=1}^{n}{T_{r}}=\frac{1}{2}[ \frac{1}{1}-\frac{1}{1+1.2} ]+\frac{1}{2}[ \frac{1}{1+1.2}-\frac{1}{1+2.3} ] $ $ +\frac{1}{2}[ \frac{1}{1+2.3}-\frac{1}{1+3.4} ]+…….+\frac{1}{2}[ \frac{1}{1+(n-1)n}-\frac{1}{1+n(n+1)} ] $ $ =\frac{1}{2}[ 1-\frac{1}{1+n(n+1)} ]=\frac{n(n+1)}{2(n^{2}+n+1)} $ . Trick: Checking for $ c $ . $ S_1=\frac{1}{3} $ and $ S_2=\frac{3}{7} $ which are given by (b).
BETA
