A) \[\frac{1}{24}n(n-1)(n+1)(3n+2)\]
B) \[\frac{{{n}^{2}}}{48}(n-1)(n-2)\]
C) \[\frac{1}{6}n(n+1)(n+2)(n+5)\]
D) None of these
Correct Answer: A
Solution :
We know that \[{{\left\{ \frac{n}{2}(n+1) \right\}}^{2}}={{(1+2+......+n)}^{2}}=\sum\limits_{1}^{n}{{{r}^{2}}}+2\sum\limits_{s<t}{st}\] \[\Rightarrow \]\[\sum\limits_{s<t}{st}=\frac{1}{2}\left\{ \frac{{{n}^{2}}{{(n+1)}^{2}}}{4}-\frac{n(n+1)(2n+1)}{6} \right\}\] \[=\frac{n}{24}(n-1)(n+1)(3n+2)\]. Trick: \[{{S}_{n}}=1\ .\ 2+2\ .\ 3+3\ .\ 4+........+(n-1)\ .\ n\] Check by putting \[(n-1)=1,\ 2\ i.e.,\ n=2,\ 3\] in the options.
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