Sequence And Series Question 532
Question: $ n^{th} $ term of the series $ \frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+…… $ will be
[Pb. CET 2000]
Options:
A) $ n^{2}+2n+1 $
B) $ \frac{n^{2}+2n+1}{8} $
C) $ \frac{n^{2}+2n+1}{4} $
D) $ \frac{n^{2}-2n+1}{4} $
Show Answer
Answer:
Correct Answer: C
Solution:
Obviously $ T_{n}=\frac{1^{3}+2^{3}+3^{3}+………+n^{3}}{1+3+5+………upto\ n\ terms} $ = $ \frac{\Sigma n^{3}}{\frac{n}{2}[2+(n-1)2]}=\frac{1}{4}\frac{n^{2}{{(n+1)}^{2}}}{n^{2}}=\frac{1}{4}(n^{2}+2n+1) $ .
BETA
