Sequence And Series Question 116
Question: The sum of $ (n-1) $ terms of $ 1+(1+3)+ $ $ (1+3+5)+……. $ is
[RPET 1999]
Options:
A) $ \frac{n,(n+1),(2n+1)}{6} $
B) $ \frac{n^{2}(n+1)}{4} $
C) $ \frac{n,(n-1),(2n-1)}{6} $
D) $ n^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ T_{n} $ be the nth term of the series $ T_{n}=2\sum{n}-\sum{1} $
Þ $ T_{n}=\frac{2n(n+1)}{2}-n=n^{2} $
$ \therefore S_{n}=\sum\limits_{k=1}^{n}{(k^{2})}=\frac{n(n+1)(2n+1)}{6} $ Hence sum of $ (n-1) $ terms $ {S_{n-1}}=\frac{(n-1),n,(2n-1)}{6} $ .
BETA
