Discussion Forum

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

Answer:

Explanation:

We have,

 $T_{r}=r[n-(r-1)]$

= r(n-r+1)

= $nr-r^{2}+r$

 $\therefore$     $\sum_{r=1}^{n}T_{r}=\sum_{r=1}^{n} (nr-r^{2}+r)$

$= \sum_{r=1}^{n}[(n+1)r-r^{2}]=(n+1)\sum_{r=1}^{n}r-\sum_{r=1}^{n}r^{2}$

  $= \frac{(n+1)n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}$

  $= \frac{n(n+1)}{2}\left[(n+1)-\left(\frac{2n+1}{3}\right)\right]$

$= \frac{n(n+1)}{2}\left[\frac{3n+3-2n-1}{3}\right]$

  $= \frac{n(n+1)}{2}\left[\frac{n+2}{3}\right]$

$= \frac{n(n+1)(n+2)}{6}$

 $\therefore$     $\alpha$  = $\frac{1}{6}$