Sequence And Series Question 68
Question: $ n^{th} $ term of the series $ 2+4+7+11+……. $ will be
[Roorkee 1977]
Options:
A) $ \frac{n^{2}+n+1}{2} $
B) $ n^{2}+n+2 $
C) $ \frac{n^{2}+n+2}{2} $
D) $ \frac{n^{2}+2n+2}{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ S=2+4+7+11+16+……..+T_{n} $ $ S=2+4+7+11+16+……..{T_{n-1}}+T_{n} $ Subtracting, we get $ 0=2+{ 2+3+4+……..+(T_{n}-{T_{n-1}}) }-T_{n} $
$ \Rightarrow $ $ T_{n}=1+(1+2+3+4+……upto\ n\ terms) $
$ \Rightarrow $ $ 1+\frac{1}{2}n(n+1)=\frac{2+n^{2}+n}{2}=\frac{n^{2}+n+2}{2} $ .
BETA
