If α∈ℝ, n∈ℕ and n+2(n-1)+3(n-2)+…(n-1)2+n.1=αn(n+1)(n+2), then α=

n+29n-10+3(n-2)+.......+(n-1) 2+n.1 =α n(n+1) (n+2)tn=r (n+1-r)Sn=∑tn=∑rn+1-r2 =(n+1) ∑r-∑r2 =(n+1) n(n+1)2-n(n+1) (2n+1)6 =n(n+1)2 n+1-2n+13 =n (n+1) (n+2)6α=16

If α∈ℝ, n∈ℕ and n+2(n-1)+3(n-2)+…(n-1)2+n.1=αn(n+1)(n+2), then α=

n+29n-10+3(n-2)+.......+(n-1) 2+n.1 =α n(n+1) (n+2)tn=r (n+1-r)Sn=∑tn=∑rn+1-r2 =(n+1) ∑r-∑r2 =(n+1) n(n+1)2-n(n+1) (2n+1)6 =n(n+1)2 n+1-2n+13 =n (n+1) (n+2)6α=16

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

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$\frac{1}{6}$

$\frac{1}{5}$

$\frac{1}{2}$

$\frac{1}{3}$

answer is D.

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Detailed Solution

$\begin{array}{rcl} n + 29 n - 10 + 3 (n - 2) + . . . . . . . + (n - 1) 2 + n . 1 \\ = & α n (n + 1) (n + 2) \\ t_{n} & = & r (n + 1 - r) \\ S_{n} & = & \sum t_{n} = \sum (r (n + 1) - r^{2}) \\ = & (n + 1) \sum r - \sum r^{2} \\ = & (n + 1) \frac{n (n + 1)}{2} - \frac{n (n + 1) (2 n + 1)}{6} \\ = & \frac{n (n + 1)}{2} [n + 1 - \frac{2 n + 1}{3}] \\ = & \frac{n (n + 1) (n + 2)}{6} \\ α & = & \frac{1}{6} \end{array}$