Question Number 179777 by luciferit last updated on 02/Nov/22
$${p}\left({x}\right)=\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+\mathrm{2}{x}\right)\left(\mathrm{1}−\mathrm{3}{x}\right)….\left(\mathrm{1}+\mathrm{14}{x}\right)\left(\mathrm{1}−\mathrm{15}{x}\right) \\ $$$$ \\ $$$${the}\:{absolute}\:{value}\:{of}\:{the}\:{coefficient}\:{of}\:{the}\:{x}^{\mathrm{2}} \:{in}\:{the}\:{expression}? \\ $$
Answered by mr W last updated on 02/Nov/22
![p(x)=Π_(k=1) ^(15) [1+(−1)^k kx] coef. of x^2 : (1/2)Σ_(k=1) ^(15) {(−1)^k kΣ_(r=1_(r≠k) ) ^(15) (−1)^r r} =(1/2)Σ_(k=1) ^(15) {(−1)^k kΣ_(r=1) ^(15) (−1)^r r−(−1)^(2k) k^2 } =(1/2){(Σ_(r=1) ^(15) (−1)^r r)^2 −Σ_(k=1) ^(15) k^2 } =(1/2){[2+4+8+...+14−(1+3+5+...+15)]^2 −((15×(15+1)×(2×15+1))/6)} =(1/2){[((7×16)/2)−((8×16)/2)]^2 −((15×16×31)/6)} =−588 ∣coef. of x^2 ∣=588](https://www.tinkutara.com/question/Q179779.png)
$${p}\left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{\mathrm{15}} {\prod}}\left[\mathrm{1}+\left(−\mathrm{1}\right)^{{k}} {kx}\right] \\ $$$${coef}.\:{of}\:{x}^{\mathrm{2}} : \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\mathrm{15}} {\sum}}\left\{\left(−\mathrm{1}\right)^{{k}} {k}\underset{\underset{{r}\neq{k}} {{r}=\mathrm{1}}} {\overset{\mathrm{15}} {\sum}}\left(−\mathrm{1}\right)^{{r}} {r}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\mathrm{15}} {\sum}}\left\{\left(−\mathrm{1}\right)^{{k}} {k}\underset{{r}=\mathrm{1}} {\overset{\mathrm{15}} {\sum}}\left(−\mathrm{1}\right)^{{r}} {r}−\left(−\mathrm{1}\right)^{\mathrm{2}{k}} {k}^{\mathrm{2}} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left(\underset{{r}=\mathrm{1}} {\overset{\mathrm{15}} {\sum}}\left(−\mathrm{1}\right)^{{r}} {r}\right)^{\mathrm{2}} −\underset{{k}=\mathrm{1}} {\overset{\mathrm{15}} {\sum}}{k}^{\mathrm{2}} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left[\mathrm{2}+\mathrm{4}+\mathrm{8}+…+\mathrm{14}−\left(\mathrm{1}+\mathrm{3}+\mathrm{5}+…+\mathrm{15}\right)\right]^{\mathrm{2}} −\frac{\mathrm{15}×\left(\mathrm{15}+\mathrm{1}\right)×\left(\mathrm{2}×\mathrm{15}+\mathrm{1}\right)}{\mathrm{6}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left[\frac{\mathrm{7}×\mathrm{16}}{\mathrm{2}}−\frac{\mathrm{8}×\mathrm{16}}{\mathrm{2}}\right]^{\mathrm{2}} −\frac{\mathrm{15}×\mathrm{16}×\mathrm{31}}{\mathrm{6}}\right\} \\ $$$$=−\mathrm{588} \\ $$$$\mid{coef}.\:{of}\:{x}^{\mathrm{2}} \mid=\mathrm{588} \\ $$