A) \[\frac{{{n}^{2}}+n}{3}\]
B) \[\frac{{{n}^{3}}+8n}{3}\]
C) \[\frac{{{n}^{2}}+8n}{5}\]
D) \[\frac{{{n}^{2}}-8n}{3}\]
Correct Answer: B
Solution :
Here, \[{{T}_{n}}=3+n(n-1)=3+{{n}^{2}}-n\] Now sum \[S=\Sigma {{T}_{n}}=\Sigma (3+{{n}^{2}}-n)\] \[=3n+\frac{1}{6}n(n+1)(2n+1)-\frac{n(n+1)}{2}\] \[=\frac{1}{6}n(n+1)[2n+1-3]+3n=\frac{{{n}^{3}}+8n}{3}\].
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