A) \[\frac{n{{(n+1)}^{3}}}{3}\]
B) \[\frac{(n+1){{n}^{2}}}{6}\]
C) \[\frac{n(n-1)(2n+1)}{12}\]
D) \[\frac{n(n+1)(2n+1)}{6}\]
Correct Answer: D
Solution :
(d): Consider the identify \[{{n}^{3}}-{{(n-1)}^{3}}-3{{n}^{2}}-3n+1\] Now, RHS \[\sum{\left( 3{{n}^{2}}-3n+1 \right)=3\times \sum{{{n}^{2}}-3\sum{n+\sum{1}}}}\] \[=3S-\frac{3n(n+1)}{2}+n\] Taking the summation of LHS \[\sum\limits_{1}^{n}{{{n}^{3}}-{{(n-1)}^{3}}={{n}^{3}}}\] \[\therefore {{n}^{3}}=3S-\frac{3n(n+1)}{2}+n\] \[\therefore 3S={{n}^{3}}+\frac{3n(n+1)}{2}-n\] \[\therefore 6S=2{{n}^{3}}+3{{n}^{2}}+3n-2n\] \[=2{{n}^{3}}+3{{n}^{2}}+n=n\,[2{{n}^{2}}+3n+1]\] \[=n\,(2n+1)\,(n+1)\] \[=\frac{n\,(n+1)\,(2n+1)}{6}\]
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