A) \[\frac{n(n+1)}{2}\]
B) \[\frac{n(n+1)(n+2)}{6}\]
C) \[\frac{n({{n}^{2}}-1)(3n+2)}{12}\]
D) \[2{{n}^{3}}+3{{n}^{2}}-1\]
Correct Answer: C
Solution :
The sum of total number of possibilities \[=1(2+3+....+n)+2(1+3+4+....+n)\] \[+....n[1+2+.....+(n-1)]\] \[=1(\Sigma n-1)+2(\Sigma n-2)+...+n(\Sigma n-n)\] \[=\Sigma n(1+2+...+n)+(-{{1}^{2}}-{{2}^{2}}+.....-{{n}^{2}})\] \[={{(\Sigma n)}^{2}}-.....(\Sigma {{n}^{2}})\] \[={{\left[ \frac{n(n+1)}{2} \right]}^{2}}-\frac{n(n+1)(2n+1)}{6}\] \[=n(n+1)\left[ \frac{n(n+1)}{4}-\frac{2n+1}{6} \right]\] \[=\frac{n(n+1)}{12}[3{{n}^{2}}+3n-.....4n-2]\] \[=\frac{n(n+1)}{12}[3{{n}^{2}}-n-2]\] \[=\frac{n(n+1)}{12}[3{{n}^{2}}-3n+2n-2]\] \[=\frac{n(n+1)}{12}[3n(n-1)+2(n-1)]\] \[=\frac{n(n+1)(3n+2)(n-1)}{12}\] \[=\frac{n({{n}^{2}}-1)(3n+2)}{12}\]
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