A) \[\frac{(n+1)(n+2)}{3}\]
B) \[n(n+1)(n+2)\]
C) \[\frac{n(n+1)(n+2)}{6}\]
D) \[\frac{n(n+1)(n+2)}{2}\]
Correct Answer: C
Solution :
[c] \[S=\sum{{{T}_{n}}=\sum{\frac{{{1}^{3}}+{{2}^{3}}+...+{{n}^{3}}}{1+2+...+n}}=\sum{\frac{\sum{{{n}^{3}}}}{\sum{n}}}}\] \[=\sum{\frac{\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}}{\frac{n(n+1)}{2}}}=\sum{\frac{n(n+1)}{2}=\frac{1}{2}[\sum{{{n}^{2}}+\sum{n}}]}\] \[=\frac{1}{2}\left[ \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} \right]\] \[=\frac{n(n+1)}{4}\left[ \frac{2n+1}{3}+1 \right]=\frac{n(n+1)(2n+4)}{12}\] \[=\frac{n(n+1)(n+2)}{6}.\]
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