Sequence And Series Question 161
Question: $ \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}}+…..n $ terms =
[EAMCET 2000]
Options:
A) $ {{( \frac{n}{n+1} )}^{2}} $
B) $ {{( \frac{n}{n+1} )}^{3}} $
C) $ ( \frac{n}{n+1} ) $
D) $ ( \frac{1}{n+1} ) $
Show Answer
Answer:
Correct Answer: C
Solution:
$ T_{n}=\frac{\frac{n(n+1)}{2.,2}}{1^{3}+2^{3}+3^{3}+…..+n^{3}}=\frac{\frac{n(n+1)}{4}}{{{( \frac{n(n+1)}{2} )}^{2}}} $ $ =\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} $ \ $ S_{n}=\sum\limits_{{}}^{{}}{( \frac{1}{n}-\frac{1}{n+1} )} $ $ =( 1-\frac{1}{2} )+( \frac{1}{2}-\frac{1}{3} )+( \frac{1}{3}-\frac{1}{4} )+…….+( \frac{1}{n}-\frac{1}{n+1} ) $ $ =1-\frac{1}{n+1}=\frac{n}{n+1} $ .
BETA
