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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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Verified Answer
The correct answer is:
$\frac{1}{6}$
We have,
$$
\begin{aligned}
T_r & =r[n-(r-1)] \\
& =r(n-r+1) \\
& =n r-r^2+r \\
\therefore \quad S_n & =\sum_{r=1}^n T_r=\sum_{r=1}^n\left(n r-r^2+r\right) \\
& =\sum_{r=1}^n\left[(n+1) r-r^2\right]=(n+1) \sum_{r=1}^n r-\sum_{r=1}^n r^2
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{(n+1) n(n+1)}{2}-\frac{n(n+1)(2 n+1)}{6} \\
& =\frac{n(n+1)}{2}\left[(n+1)-\left(\frac{2 n+1}{3}\right)\right] \\
& =\frac{n(n+1)}{2}\left[\frac{3 n+3-2 n-1}{3}\right] \\
& =\frac{n(n+1)}{2}\left[\frac{n+2}{3}\right] \\
& =\frac{n(n+1)(n+2)}{6} \\
\therefore \quad \alpha & =\frac{1}{6}
\end{aligned}
$$
$$
\begin{aligned}
T_r & =r[n-(r-1)] \\
& =r(n-r+1) \\
& =n r-r^2+r \\
\therefore \quad S_n & =\sum_{r=1}^n T_r=\sum_{r=1}^n\left(n r-r^2+r\right) \\
& =\sum_{r=1}^n\left[(n+1) r-r^2\right]=(n+1) \sum_{r=1}^n r-\sum_{r=1}^n r^2
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{(n+1) n(n+1)}{2}-\frac{n(n+1)(2 n+1)}{6} \\
& =\frac{n(n+1)}{2}\left[(n+1)-\left(\frac{2 n+1}{3}\right)\right] \\
& =\frac{n(n+1)}{2}\left[\frac{3 n+3-2 n-1}{3}\right] \\
& =\frac{n(n+1)}{2}\left[\frac{n+2}{3}\right] \\
& =\frac{n(n+1)(n+2)}{6} \\
\therefore \quad \alpha & =\frac{1}{6}
\end{aligned}
$$
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