A) \[{{n}^{3}}+{{n}^{2}}+n+2\]
B) \[\frac{1}{6}(2{{n}^{3}}+12{{n}^{2}}+10n-84)\]
C) \[{{n}^{3}}+{{n}^{2}}+n\]
D) None of these
Correct Answer: B
Solution :
\[S=3\ .\ 6+4\ .\ 7+.......\]upto \[n-2\] 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 \[n=3,\ 4,\ 5.......\] Trick: \[{{S}_{1}}=18,\ {{S}_{2}}=46\] Now put in options \[(n-2)=1,\ 2\ \ i.e.\ \ n=3,\ 4\] Obviously (b) gives the values.
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