If alpha in R, n in N and n+2(n-1)+3(n-2)+ldo | vedclass

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(D) The general term of the series is $T_r = r(n-r+1) = nr - r^2 + r$. Summing from $r=1$ to $n$: $S_n = \sum_{r=1}^n (nr - r^2 + r) = (n+1) \sum_{r=1

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$\frac{1}{6}$

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(D) The general term of the series is $T_r = r(n-r+1) = nr - r^2 + r$. Summing from $r=1$ to $n$: $S_n = \sum_{r=1}^n (nr - r^2 + r) = (n+1) \sum_{r=1}^n r - \sum_{r=1}^n r^2$. Using standard summation formulas: $S_n = (n+1) \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$. Factoring out $\frac{n(n+1)}{2}$: $S_n = \frac{n(n+1)}{2} [ (n+1) - \frac{2n+1}{3} ] = \frac{n(n+1)}{2} [ \frac{3n+3-2n-1}{3} ] = \frac{n(n+1)(n+2)}{6}$. Comparing this with $\alpha n(n+1)(n+2)$,we get $\alpha = \frac{1}{6}$.

If $\alpha \in R, n \in N$ and $n+2(n-1)+3(n-2)+\ldots+(n-1)2+n.1 = \alpha n(n+1)(n+2)$,then $\alpha =$

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