Sequence And Series Question 149
Question: The sum of $ n $ terms of the following series $ 1.2+2.3+3.4+4.5+……… $ shall be
[MNR 1980]
Options:
A) $ n^{3} $
B) $ \frac{1}{3}n,(n+1)(n+2) $
C) $ \frac{1}{6}n,(n+1)(n+2) $
D) $ \frac{1}{3}n,(n+1)(2n+1) $
Show Answer
Answer:
Correct Answer: B
Solution:
The first factors of the terms of the given series $ 1,\ 2,\ 3,\ 4,\ ……..n $ and second factors of the terms of the given series $ 2,\ 3,\ 4,,\ ……..(n+1) $ $ n^{th} $ term of the given series $ =n(n+1)=n^{2}+n $ Hence sum = $ \Sigma n^{2}+\Sigma n=\frac{1}{6}n(n+1)(2n+1)+\frac{n}{2}(n+1) $ $ =\frac{1}{6}n(n+1)(2n+1+3)=\frac{1}{3}n(n+1)(n+2) $ .
BETA
