Coefficient of $x^2$

Let $y$ be the polynomial. Then $\begin{aligned} \ln(y) &= \ln(1-x) + \ln(1+2x) + \cdots + \ln(1-15x) \\ \frac{y'}{y} &= -\frac1{1-x} + \frac2{1+2x} - \cdots - \frac{15}{1-15x}. \end{aligned}$ Plugging in $x=0$ gives $y=1$ and $y' = -1+2-3+\cdots+14-15 = -8.$

Taking one more derivative gives $\frac{y(y'')-(y')^2}{y^2} = -\frac1{(1-x)^2} - \frac4{(1+2x)^2} - \cdots - \frac{225}{(1-15x)^2},$ so plugging in $x=0$ as before gives $y''-(-8)^2 = -1-4-\cdots-225,$ so $y''(0) = 64 - \sum_{k=1}^{15} k^2 = 64 - \frac{15(16)(31)}6 = 64 - 1240 = -1176.$ Now $y''(0)$ is just $2C,$ so $C = -588,$ and the answer is $\fbox{588}.$

To get $x^{2}$ you multiply $x$ terms from two factors and $1s$ from other factors. Basically, we need to find all combinations of these $x$ terms, calculate and sum them. There are $\binom{15}{2} = 105$ of them, but I decided to group them nicely according to the first term: all combinations with $-x$ , then $2x$ , $-3x$ , and so on. Then I wrote a python code to do that faster for me, I hope it's understandable.

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import numpy as np

n = np.array(range(1, 16)) #creating an array consisted of numbers 1 to 15
print(n)

for i in range(15): #editing an array so that every odd number becomes negative to match x-term coefficients from the problem
    if n[i] % 2 == 0:
        n[i] = n[i]
    else:
        n[i] = -n[i]
print(n)              

sum = np.zeros((15)) #initializing array "sum"
for i in range(14): 
    sum[i] = np.sum(n[i+1:15]) * n[i] #calculating sum for every "group"
print(sum)

whole_sum = np.sum(sum)
print(whole_sum)

Let given expression 'E' $coffecient\quad of\quad { x }^{ 2 }\quad =\quad \frac { 1 }{ 2 } \quad lim\quad x\longrightarrow 0\quad \{ { D }^{ '' }(E)\} =588\\ where\quad { D }^{ '' }(E)\quad means\quad double\quad derivative\quad of\quad E\quad with\quad respect\quad to\quad x\\ \quad$ . NOTE: use logarithmic differentiation.(for easy calculation)

Here is a non-calculus/non-programming partial solution.

I will give a recursive forumula without proof:

$F_n = F_{n-1} - n\lfloor\dfrac{n}{2}\rfloor$

By evaluating the first two products as $(1-x)(1+2x)$ , the coefficient of $x^2$ is $-2$ . So let $F_2 = -2$ .

Applying the recursive formula until $F_{15}$ we get $\boxed{F_{15} = 588}$ .

Mathematica:

Abs@Coefficient[Product[1 + (-1)^n*n*x, {n, 15}], x^2]

588

Let $f(x) = (1-x)(1+2x)(1-3x)(1+4x)\cdots(1+14x)(1-15x)$ . We need to find $\dfrac {f''(0)}{2!} = C$ and we have:

$\begin{aligned} f(x) & = \prod_{k=1}^{15} \left(1+(-1)^kkx\right) \\ \implies f(0) & = 1 \\ f'(x) & = \sum_{k=1}^{15} \frac {(-1)^kkf(x)}{1+(-1)^k kx} \\ \implies f'(0) & = \sum_{k=1}^{15} \frac {(-1)^kkf(0)}{1+(-1)^k k(0)} = \sum_{k=1}^{15} (-1)^kk \\ & = {\color{#3D99F6}-1+2}{\color{#D61F06}-3+4}{\color{#3D99F6}-5+6} \cdots {\color{#3D99F6}-13+14}\color{#D61F06}-15 \\ & = {\color{#3D99F6}1}+{\color{#D61F06}1} + {\color{#3D99F6}1}+{\color{#D61F06}1}+{\color{#3D99F6}1}+{\color{#D61F06}1}+{\color{#3D99F6}1}{\color{#D61F06}-15} \\ & = - 8 \\ f''(x) & = \sum_{k=1}^{15} \frac {(-1)^kk\left(f'(x)\left(1+(-1)^k kx\right)-(-1)^kkf(x)\right)}{\left(1+(-1)^k kx\right)^2} \\ \implies f''(0) & = \sum_{k=1}^{15} \left((-1)^kk(-8) - k^2\right) \\ & = -8 \sum_{k=1}^{15} (-1)^k k - \sum_{k=1}^{15} k^2 \\ & = -8 (-8) - \frac {15(16)(31)}6 \\ & = -1176 \\ \implies C & = - \frac {1176}2 = - 588 \\ |C| & = \boxed{588} \end{aligned}$

$\color{#EC7300}{Let~a_n=1+(-1)^n*n*x.\\ Make~a~note:-~~~In~a_n*a_{n+1}~ there~are~{ n*(n+1)} ~~-~x^2s.~~~~and~n~is~an~integer.\\ \therefore~In~~a_n*a_{n+1}+a_n*a_{n+2}=a_n*\{a_{n+1}*a_{n+2}\}~there~is~only~0NE~~+x^2.\\ So~for~each~a_n,~we~can~make~a~pair~of~two~adjoining~links,~a_r~and~a_{r+1}~,~n<r<16,\\ and~whose~sum~will~have~only~one~+x^2.\\ But ~for~ even~n~after~pairing~~-~a_{15}~remains,~so~that~ will~ get ~~~~n*15~~-~x^2s~~in~it.\\ So~out~of~the~parings,~we~get~\displaystyle~\sum_{n=1}^{15} n*\lfloor \dfrac{15-n} 2 \rfloor=\color{#3D99F6}{252 ~of~+x^2s}.\\ In~even~n=2m~~with~a_{15}~~we~get~~~~\displaystyle~\sum_{m=1}^{7} 2*m*( - 15)=\dfrac{7*(7+1)}2*2*( - 15) =\color{#3D99F6}{840 ~of~~-~x^2s}.\\ So~|C|=|+252-840|=\Large \color{#D61F06}{840-252=588} }\\ ~~~~~\\ Explanation:- \\ \color{#3D99F6}{Take~n=6.~~(1+6x)(1-7x)(1+8x)(1-9x)(1+10x)(1-11x)(1+12x)(1-13x)(1+14x)(1-15x) \\ For~x^2:-(1+6x)* \Bigg \{(1\!-7x)+(1+8x)~~+~(1\!-9x)+(1+10x)~~+~(1\!-11x)+(1+12x)~~+~(1\!-13x)+(1+14x)~~+~(1-15x) \Bigg \} \\ .......\implies~ ~+6x~~* \Bigg \{~-~7x+8x~~~+~~~~-9x+10x~~+~~~~~-~11x+12x~~+~~~~~~~~~~-~13x+14x~~~+~~~~~~~~-15x) \Bigg \} \\ ....................~+6x~~*~~\Bigg \{~~~~+1x~~~~~~~~~~+1x~~~~~~~~~~~~~~+1x~~~~~~~~~~~~~~~~~+1~~~~~~-15x) \Bigg \} \\ ......................\Bigg \{+6*(1+1+1+1)~~~~~~~~~~~~~~~~~~~~~+6*(-15) \Bigg \}*x^2 .\\ \color{#D61F06}{(24~~~~~~~~~~~~~~~~~~~~-90)*x^2} .\\ ~~~~By~our~formula~above:-\\ ..n=2m...................(2m*\lfloor \dfrac{15-n} 2 \rfloor~~~~~~~~ n*(- 15))*x^2 \\ ......................\{ 6*\lfloor \dfrac{15-6} 2 \rfloor~~~~~~~~~~6*(- 15)\}*x^2 \\ \color{#D61F06}{(6*4~~~~~~~~~~~~~~~~~~~~~~~~~~~~- 90)*x^2.} }$