A) \[(n+1)\,(n+2)\,(n+3)/6\]
B) \[n\,(n+1)\,(n+2)/6\]
C) \[n\,(n+1)\,(2n+3)\,\]
D) \[n\,(n+1)\,(2n+1)/6\]
Correct Answer: D
Solution :
\[S=(2n-1)+2(2n-3)+3(2n-5)+....\] \[S=[2n+2.2n+3.2n+......+n.2n]-\]\[[1+2.3+3.5+....+n.(2n-1)]\] Let,\[{{S}_{1}}=2n(1+2+3+....+n)\]=\[\frac{2n.n(n+1)}{2}={{n}^{2}}(n+1)\] and \[{{S}_{2}}=1+2.3+3.5+.....+n.(2n-1)\] \[{{T}_{n}}=n(2n-1)=2{{n}^{2}}-n\] \[\therefore {{S}_{2}}=\sum (2{{n}^{2}}-n)\]\[=2\sum ({{n}^{2}})-\sum (n)\] \[=\frac{2n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}\] \[\therefore S={{S}_{1}}-{{S}_{2}}={{n}^{2}}(n+1)-\frac{2n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\] \[=n\,(n+1)\left[ n-\frac{2n+1}{3}+\frac{1}{2} \right]\] \[=n\,(n+1)\left[ \frac{6n-4n-2+3}{6} \right]\]\[=\frac{n\,(n+1)(2n+1)}{6}\].
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