Sequence And Series Question 141
Question: The sum of the series $ {{1.3}^{2}}+{{2.5}^{2}}+{{3.7}^{2}}+………. $ upto $ 20 $ terms is
[IIT 1973]
Options:
A) 188090
B) 189080
C) 199080
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Here $ T_{n} $ of the A.P. $ 1,\ 2,\ 3,\ ……….=n $ and $ T_{n} $ of the A.P. $ 3,\ 5,\ 7,……..=2n+1 $
$ \therefore $ $ T_{n} $ of given series $ =n{{(2n+1)}^{2}}=4n^{3}+4n^{2}+n $ Hence $ S=\sum\limits_1^{20}{T_{n}}=4\sum\limits_1^{20}{n^{3}}+4\sum\limits_1^{20}{n^{2}}+\sum\limits_1^{20}{n} $ $ =4\cdot \frac{1}{4}20^{2}\cdot 21^{2}+4\cdot \frac{1}{6}20\cdot 21\cdot 41+\frac{1}{2}20\cdot 21=188090 $
BETA
