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Question Number 85365 by Power last updated on 21/Mar/20

	Answered by jagoll last updated on 21/Mar/20

	$C = S_{112} = \frac{112}{2} (- 4 + (111 \times (- 8))$ $= 56 \times (- 892)$ $= - 49952$
	Commented by Power last updated on 21/Mar/20

	$?$
	Commented by john santu last updated on 21/Mar/20

	$c o n s i d e r (1 - x) (1 + 2 x) =$ $(1 - x) (1 + x + x) = 1 - x^{2} + (1 - x) x$ $= 1 - x^{2} + x - x^{2} = 1 + x - 2 x^{2}$ $(1 - 3 x) (1 + 4 x) = (1 - 3 x) (1 + 3 x + x)$ $= 1 - 9 x^{2} + x (1 - 3 x)$ $= 1 + x - 12 x^{2}$ $(1 - 5 x) (1 + 6 x) = (1 - 5 x) (1 + 5 x + x)$ $= 1 - 25 x^{2} + x (1 - 5 x)$ $= 1 + x - 30 x^{2}$ $n o w : (1 + x - 2 x^{2}) (1 + x - 12 x^{2}) (1 + x - 30 x^{2}) \times . . .$ $\times (1 + x - 892 x^{2})$
	Commented by Power last updated on 21/Mar/20

	$i do not understand$
	Commented by john santu last updated on 21/Mar/20

	$o u w h$
	Answered by mr W last updated on 21/Mar/20
	$(1−(2n−1)x)(1+2nx)=1+x−2n(2n−1)x^2 {(1−x)(1+2x)(1−3x)(1+4x)...(1−221x)(1+222x)}(1−223x) ={Π_(n=1) ^(111) [1−(2n−1)x](1+2nx)}(1−223x) ={Π_(n=1) ^(111) [1+x−2n(2n−1)x^2 ]}(1−223x) coef. of x^2 : −Σ_(n=1) ^(111) 2n(2n−1)+((111×110)/2)−223×111 =−Σ_(n=1) ^(111) (4n^2 −2n)+((111×110)/2)−223×111 =−(4×((111×112×223)/6)−2×((111×112)/2))+((111×110)/2)−223×111 =−4×((111×112×223)/6)−56×111 =−1 854 440 ⇒ answer$
	$(1 - (2 n - 1) x) (1 + 2 n x) = 1 + x - 2 n (2 n - 1) x^{2}$ ${(1 - x) (1 + 2 x) (1 - 3 x) (1 + 4 x) . . . (1 - 221 x) (1 + 222 x)} (1 - 223 x)$ $= {\underset{n = 1}{\prod^{111}} [1 - (2 n - 1) x] (1 + 2 n x)} (1 - 223 x)$ $= {\underset{n = 1}{\prod^{111}} [1 + x - 2 n (2 n - 1) x^{2}]} (1 - 223 x)$ $c o e f . o f x^{2} :$ $- \underset{n = 1}{\sum^{111}} 2 n (2 n - 1) + \frac{111 \times 110}{2} - 223 \times 111$ $= - \underset{n = 1}{\sum^{111}} (4 n^{2} - 2 n) + \frac{111 \times 110}{2} - 223 \times 111$ $= - (4 \times \frac{111 \times 112 \times 223}{6} - 2 \times \frac{111 \times 112}{2}) + \frac{111 \times 110}{2} - 223 \times 111$ $= - 4 \times \frac{111 \times 112 \times 223}{6} - 56 \times 111$ $= - 1 854 440 \Rightarrow a n s w e r$
	Commented by Power last updated on 21/Mar/20

	$thanks$
	Answered by mr W last updated on 21/Mar/20
	$an other way (1−x)(1+2x)(1−3x)(1+4x)...(1−223x) =(1+a_2 x)(1+a_2 x)(1+a_3 x)...(1+a_(223) x) with a_n =(−1)^n n coef. of x^2 : C_2 =Σ_(i,j=1,2,...,223_(j>i) ) a_i a_j =(1/2){Σ_(i=1) ^(223) a_i (Σ_(j=1) ^(223) a_j −a_i )} =(1/2){(Σ_(i=1) ^(223) a_i )^2 −Σ_(i=1) ^(223) a_i ^2 } Σ_(i=1) ^(223) a_i =(−1+2)+(−3+4)−...+(−221+222)−223 =1×111−223 =112 Σ_(i=1) ^(223) a_i ^2 =(−1)^2 +2^2 +(−3)^2 +...+(−223)^2 =((223×224×(2×223+1))/6)=3 721 424 ⇒C_2 =(1/2)(112^2 −3 721 424)=−1 854 440$
	$a n o t h e r w a y$ $(1 - x) (1 + 2 x) (1 - 3 x) (1 + 4 x) . . . (1 - 223 x)$ $= (1 + a_{2} x) (1 + a_{2} x) (1 + a_{3} x) . . . (1 + a_{223} x)$ $w i t h a_{n} = {(- 1)}^{n} n$ $c o e f . o f x^{2} :$ $C_{2} = \sum_{\underset{j > i}{i, j = 1, 2, . . ., 223}} a_{i} a_{j}$ $= \frac{1}{2} {\underset{i = 1}{\sum^{223}} a_{i} (\underset{j = 1}{\sum^{223}} a_{j} - a_{i})}$ $= \frac{1}{2} {{(\underset{i = 1}{\sum^{223}} a_{i})}^{2} - \underset{i = 1}{\sum^{223}} a_{i}^{2}}$ $\underset{i = 1}{\sum^{223}} a_{i} = (- 1 + 2) + (- 3 + 4) - . . . + (- 221 + 222) - 223$ $= 1 \times 111 - 223$ $= 112$ $\underset{i = 1}{\sum^{223}} a_{i}^{2} = {(- 1)}^{2} + 2^{2} + {(- 3)}^{2} + . . . + {(- 223)}^{2}$ $= \frac{223 \times 224 \times (2 \times 223 + 1)}{6} = 3 721 424$ $\Rightarrow C_{2} = \frac{1}{2} (112^{2} - 3 721 424) = - 1 854 440$
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