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

$$\mathrm{The}\mathrm{coefficient}\mathrm{of}{x}^{n-2}\mathrm{in}\mathrm{the}\mathrm{polynomial}$$$$(x-1)(x-2)....(x-n)\mathrm{is}$$

Commented by Yozzia last updated on 01/Oct/16

$$coefficient\left(x^{n-2}\right)=\sum _{\mathrm{\forall }\alpha ,\beta \in [1,n]}\alpha \beta$$$$1\times 2+1\times 3+1\times 4+...+1\times n$$$$+2\times 3+2\times 4+2\times 5+...+2\times n$$$$+3\times 4+3\times 5+...+3+n$$$$+...+(n-1)\times n$$$$$$$$$$

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

$$\mathrm{From}{\mathrm{Yozzi}}^{\prime }\mathrm{s}\mathrm{comment}$$$$\underset{j=1}{\sum ^{n-1}}\underset{i=j+1}{\sum ^n}j\cdot i$$$$\mathrm{Please}\mathrm{see}\mathrm{post}\mathrm{from}\mathrm{sou}Q8139\mathrm{for}\mathrm{correct}\mathrm{answer}.$$

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

$$\underset{j=1}{\sum ^{n-1}}j(\underset{i=1}{\sum ^n}i-\underset{i=1}{\sum ^j}j)$$$$=\underset{j=1}{\sum ^{n-1}}j\frac{n(n+1)}{2}-\underset{j=1}{\sum ^{n-1}}j\frac{j(j+1)}{2}$$$$=\frac{n(n+1)n(n-1)}{4}-\underset{j=1}{\sum ^{n-1}}\frac{j^3+j^2}{2}$$$$=\frac{n^2(n-1)(n+1)}{4}-\frac{1}{2}{\left\{\frac{(n-1)n}{2}\right\}}^2-\frac{1}{2}\left\{\frac{(n-1)n(2n-1)}{6}\right\}$$$$=\frac{n(n-1)}{24}[6n(n+1)-3n(n-1)-2(2n-1)]$$$$=\frac{n(n-1)}{24}[6n^2+6n-3n^2+3n-4n+2]$$$$=\frac{n(n-1)}{24}[3n^2+5n+2]$$$$=\frac{n(n-1)}{24}[3n^2+3n+2n+2]$$$$=\frac{n(n-1)(n+1)(3n+2)}{24}$$

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