Question: Sum of   $ \frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+2}+… $   to n terms is
Options:
A)   $ \frac{(n+1)(n+2)}{3} $
B)   $ n(n+1)(n+2) $
C)   $ \frac{n(n+1)(n+2)}{6} $
D)   $ \frac{n(n+1)(n+2)}{2} $
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  Answer:
Correct Answer: C
Solution:
- [c]   $ S=\sum{T_{n}=\sum{\frac{1^{3}+2^{3}+…+n^{3}}{1+2+…+n}}=\sum{\frac{\sum{n^{3}}}{\sum{n}}}} $      $ =\sum{\frac{\frac{n^{2}{{(n+1)}^{2}}}{4}}{\frac{n(n+1)}{2}}}=\sum{\frac{n(n+1)}{2}=\frac{1}{2}[\sum{n^{2}+\sum{n}}]} $      $ =\frac{1}{2}[ \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} ] $      $ =\frac{n(n+1)}{4}[ \frac{2n+1}{3}+1 ]=\frac{n(n+1)(2n+4)}{12} $      $ =\frac{n(n+1)(n+2)}{6}. $