Sequence And Series Question 363
Question: $ 2+4+7+11+16+…… $ to $ n $ terms =
[Roorkee 1977]
Options:
A) $ \frac{1}{6}(n^{2}+3n+8) $
B) $ \frac{n}{6}(n^{2}+3n+8) $
C) $ \frac{1}{6}(n^{2}-3n+8) $
D) $ \frac{n}{6}(n^{2}-3n+8) $
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ S=2+4+7+11+16+…..+T_{n} $ Again $ S=\text{ }2+4+7+11+…….+{T_{n-1}}+T_{n} $ Subtracting, we get $ 0=2+{ 2+3+4+5+…..(T_{n}-{T_{n-1}}) }-T_{n} $ $ T_{n}=2+\frac{1}{2}(n-1)(4+{n-2)1}=\frac{1}{2}(n^{2}+n+2) $ Now $ S=\Sigma T_{n}=\frac{1}{2}\Sigma (n^{2}+n+2)=\frac{1}{2}(\Sigma n^{2}+\Sigma n+2\Sigma ,1) $ $ =\frac{1}{2}{ \frac{1}{6}n(n+1)(2n+1)+\frac{1}{2}n(n+1)+2n } $ $ =\frac{n}{12}{ (n+1)(2n+1+3)+12 } $ = $ \frac{n}{6}{ (n+1)(n+2)+6 }=\frac{n}{6}(n^{2}+3n+8) $ .
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