Sequence And Series Question 151
Question: The sum to $ n $ terms of the infinite series $ {{1.3}^{2}}+{{2.5}^{2}}+{{3.7}^{2}}+……….\infty $ is
[AMU 1982]
Options:
A) $ \frac{n}{6}(n+1)(6n^{2}+14n+7) $
B) $ \frac{n}{6}(n+1)(2n+1)(3n+1) $
C) $ 4n^{3}+4n^{2}+n $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
This is an A.G. series whose $ n^{th} $ term is equal to $ T_{n}=n{{(2n+1)}^{2}}=4n^{3}+4n^{2}+n $
$ \therefore $ $ S_{n}=\sum\limits_1^{n}{T_{n}}=\sum\limits_1^{n}{(4n^{3}+4n^{2}+n)} $ $ =4\sum\limits_1^{n}{n^{3}}+4\sum\limits_1^{n}{n^{2}}+\sum\limits_1^{n}{n} $ $ =4{{{ \frac{n}{2}(n+1) }}^{2}}+\frac{4}{6}n(n+1)(2n+1)+\frac{n}{2}(n+1) $ $ =\frac{n}{6}(n+1)(6n^{2}+14n+7) $ .
BETA
