p(x)=(1−x)(1+2x)(1−3x)....(1+14x)(1−15x) the absolute value of the coefficient of the x^2 in the expression?


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Question Number 179777 by luciferit last updated on 02/Nov/22

$p (x) = (1 - x) (1 + 2 x) (1 - 3 x) . . . . (1 + 14 x) (1 - 15 x)$ $t h e a b s o l u t e v a l u e o f t h e c o e f f i c i e n t o f t h e x^{2} i n t h e e x p r e s s i o n ?$
	Answered by mr W last updated on 02/Nov/22
	$p(x)=Π_(k=1) ^(15) [1+(−1)^k kx] coef. of x^2 : (1/2)Σ_(k=1) ^(15) {(−1)^k kΣ_(r=1_(r≠k) ) ^(15) (−1)^r r} =(1/2)Σ_(k=1) ^(15) {(−1)^k kΣ_(r=1) ^(15) (−1)^r r−(−1)^(2k) k^2 } =(1/2){(Σ_(r=1) ^(15) (−1)^r r)^2 −Σ_(k=1) ^(15) k^2 } =(1/2){[2+4+8+...+14−(1+3+5+...+15)]^2 −((15×(15+1)×(2×15+1))/6)} =(1/2){[((7×16)/2)−((8×16)/2)]^2 −((15×16×31)/6)} =−588 ∣coef. of x^2 ∣=588$
	$p (x) = \underset{k = 1}{\prod^{15}} [1 + {(- 1)}^{k} k x]$ $c o e f . o f x^{2} :$ $\frac{1}{2} \underset{k = 1}{\sum^{15}} {{(- 1)}^{k} k \underset{\underset{r \neq k}{r = 1}}{\sum^{15}} {(- 1)}^{r} r}$ $= \frac{1}{2} \underset{k = 1}{\sum^{15}} {{(- 1)}^{k} k \underset{r = 1}{\sum^{15}} {(- 1)}^{r} r - {(- 1)}^{2 k} k^{2}}$ $= \frac{1}{2} {{(\underset{r = 1}{\sum^{15}} {(- 1)}^{r} r)}^{2} - \underset{k = 1}{\sum^{15}} k^{2}}$ $= \frac{1}{2} {{[2 + 4 + 8 + . . . + 14 - (1 + 3 + 5 + . . . + 15)]}^{2} - \frac{15 \times (15 + 1) \times (2 \times 15 + 1)}{6}}$ $= \frac{1}{2} {{[\frac{7 \times 16}{2} - \frac{8 \times 16}{2}]}^{2} - \frac{15 \times 16 \times 31}{6}}$ $= - 588$ $∣ c o e f . o f x^{2} ∣= 588$
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