The-coefficient-of-x-n-2-in-the-polynomial-x-1-x-2-x-n-is-

Question Number 8132 by 314159 last updated on 01/Oct/16

The coefficient of x^{n - 2} in the polynomial

(x - 1) (x - 2) \dots . (x - n) is

Commented by Yozzia last updated on 01/Oct/16

c o e f f i c i e n t (x^{n - 2}) = \sum_{\forall α, β \in [1, n]} α β

1 \times 2 + 1 \times 3 + 1 \times 4 + \dots + 1 \times n

+ 2 \times 3 + 2 \times 4 + 2 \times 5 + \dots + 2 \times n

+ 3 \times 4 + 3 \times 5 + \dots + 3 + n

+ \dots + (n - 1) \times n

Answered by prakash jain last updated on 01/Oct/16

From {Yozzi}^{'} s comment

\underset{j = 1}{\sum^{n - 1}} \underset{i = j + 1}{\sum^{n}} j \cdot i

Please see post from sou Q 8139 for correct answer .

Commented by prakash jain last updated on 02/Oct/16

\underset{j = 1}{\sum^{n - 1}} j (\underset{i = 1}{\sum^{n}} i - \underset{i = 1}{\sum^{j}} j)

= \underset{j = 1}{\sum^{n - 1}} j \frac{n (n + 1)}{2} - \underset{j = 1}{\sum^{n - 1}} j \frac{j (j + 1)}{2}

= \frac{n (n + 1) n (n - 1)}{4} - \underset{j = 1}{\sum^{n - 1}} \frac{j^{3} + j^{2}}{2}

= \frac{n^{2} (n - 1) (n + 1)}{4} - \frac{1}{2} {\frac{(n - 1) n}{2}}^{2} - \frac{1}{2} {\frac{(n - 1) n (2 n - 1)}{6}}

= \frac{n (n - 1)}{24} [6 n (n + 1) - 3 n (n - 1) - 2 (2 n - 1)]

= \frac{n (n - 1)}{24} [6 n^{2} + 6 n - 3 n^{2} + 3 n - 4 n + 2]

= \frac{n (n - 1)}{24} [3 n^{2} + 5 n + 2]

= \frac{n (n - 1)}{24} [3 n^{2} + 3 n + 2 n + 2]

= \frac{n (n - 1) (n + 1) (3 n + 2)}{24}