A) \[{{n}^{3}}+{{n}^{2}}+n+2\]
B) \[\frac{1}{6}(2{{n}^{3}}+12{{n}^{2}}+10\,n-84)\]
C) \[{{n}^{3}}+{{n}^{2}}+n\]
D) None of these
Correct Answer: B
Solution :
[b] \[S=3.6+4.7+.....\]up to \[n-2\] terms \[=(1.4+2.5+3.6+4.7+.........upto\,\,n\,\,terms)-14\]\[=\Sigma n\,(n+3)-14=\frac{1}{6}(2{{n}^{3}}+12{{n}^{2}}+10n)-14\] \[=\left( \frac{2{{n}^{3}}+12{{n}^{2}}+10n-84}{6} \right),\] where \[n=3,4,5......\]
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