Sequence And Series Question 439
Question: The sum of $ \frac{\frac{1}{2}.\frac{2}{2}}{1^{3}}+\frac{\frac{2}{2}.\frac{3}{2}}{1^{3}+2^{3}}+\frac{\frac{3}{2}.\frac{4}{2}}{1^{3}+2^{3}+3^{3}}+….. $ upto n terms is equal to
Options:
A) $ \frac{n-1}{n} $
B) $ \frac{n}{n+1} $
C) $ \frac{n+1}{n+2} $
D) $ \frac{n+1}{n} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The general term is $ T_{n}=\frac{\frac{n}{2}.\frac{n+1}{2}}{1^{3}+2^{3}+3^{3}+….+n^{3}}=\frac{1}{n(n+1)} $ $ =\frac{1}{n}-\frac{1}{n+1} $
$ \therefore S_{n}=1-\frac{1}{n+1}=\frac{n}{n+1} $
BETA
