A) \[\frac{n\,({{n}^{2}}+2)\,\,(3n+1)}{24}\]
B) \[\frac{n\,({{n}^{2}}-1)\,\,(3n+2)}{24}\]
C) \[\frac{n\,({{n}^{2}}+1)\,\,(3n+4)}{24}\]
D) none of these
Correct Answer: B
Solution :
| \[E=(x-{{\alpha }_{1}})\,\,(x-{{\alpha }_{2}})\,\,(x-{{\alpha }_{3}}).....(x-{{\alpha }_{n}})\] where \[{{\alpha }_{1}}=1,\]\[{{\alpha }_{2}}=2\] etc |
| \[={{x}^{n}}-(\Sigma {{\alpha }_{1}})\,{{x}^{n\,-1}}+(\Sigma {{\alpha }_{1}}{{\alpha }_{2}})\,{{x}^{n\,-2}}......\] |
| Hence co-efficient of \[{{x}^{n\,-\,2}}=\text{sum}\]sum of the product of the first 'n' natural numbers taken two at a time. |
| Now consider \[=\frac{{{(1+2+3.....+n)}^{2}}-({{1}^{2}}+{{2}^{2}}+{{3}^{2}}.....+{{n}^{2}})}{2}\] |
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