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Question Number 177822 by cortano1 last updated on 09/Oct/22

$$\mathrm{what}\mathrm{is}\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2\mathrm{in}$$$$(1-\mathrm{x})(1+2\mathrm{x})(1-3\mathrm{x})(1+4\mathrm{x})(1-5\mathrm{x})$$$$...(1+14\mathrm{x})(1-15\mathrm{x})?$$

Commented by mr W last updated on 10/Oct/22

$$doyouhavethesolution?$$

Commented by cortano1 last updated on 11/Oct/22

$$\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2\mathrm{in}\mathrm{the}\mathrm{expansion}$$$$\mathrm{of}\mathrm{the}\mathrm{product}\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}}}(1+{\mathrm{a}}_{\mathrm{k}}\mathrm{x})$$$$\mathrm{is}\mathrm{equal}\mathrm{to}\sum _{\mathrm{i}\ne \mathrm{j}}{\mathrm{a}}_{\mathrm{i}}{\mathrm{a}}_{\mathrm{j}}$$$$\Rightarrow \sum _{\mathrm{i}\ne \mathrm{j}}{\mathrm{a}}_{\mathrm{i}}{\mathrm{a}}_{\mathrm{j}}=\frac{1}{2}(\sum _{\mathrm{i},\mathrm{j}}{\mathrm{a}}_{\mathrm{i}}{\mathrm{a}}_{\mathrm{j}}-\sum _{\mathrm{j}}{\mathrm{a}}_{\mathrm{j}}^2)$$$$=\frac{1}{2}({\left(\sum _{\mathrm{j}}{\mathrm{a}}_{\mathrm{j}}\right)}^2-\sum _{\mathrm{j}}{\mathrm{a}}_{\mathrm{j}}^2)$$$$=\frac{1}{2}({(-8)}^2-1240)$$$$=-588$$

Answered by mr W last updated on 09/Oct/22

$${(1+3+5+...+15)}^2/2+{(2+4+6+...+14)}^2/2$$$$-(1^2+2^2+3^2+...+{15}^2)/2$$$$-(1+3+5+...+15)(2+4+6+...+14)$$$$={64}^2/2+{56}^2/2-\frac{15\times 16\times (2\times 15+1)}{12}-64\times 56$$$$=-588$$

Commented by Tawa11 last updated on 10/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by cortano1 last updated on 11/Oct/22

$$\mathrm{yes}-588$$

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