Coefficient of $x^2$ in the polynomial

Question

How to find the coefficient of $x^2$ in the polynomial $$(1-x)(1+2x)(1-3x)....(1+14x)(1-15x)$$

Is there any particular way to look at such problems?

Answer 1

Since the constant term in each binomial factor is $1$, then we can express the coefficient of the $x^2$ term as the sum of the following series:

$-2+3-4+5-6+7...+15$

$-2-6+8-10+12-14...-30$

$...$

$15-30+45-60+75-90...-210$

The sum of all these sums is the answer, because for each factor, you're multiplying the $x$ term with the $x$ term from another factor and then just multiplying by the $1$ for the rest of terms to yield a $x^2$ term.

Answer 2

The terms $x^2$ appear as the product of $x$'s from two factors, while the $1$'s are used from the remaining factors.

Hence taking all pairs of factors,

$$C_2=\sum_{i=1}^{14}\sum_{j=i+1}^{15}(-1)^ii(-1)^jj.$$

You can ease this computation by taking all pairs twice, including the diagonal ones and

$$2C_2=\sum_{i=1}^{15}\sum_{j=1}^{15}(-1)^ii(-1)^jj-\sum_{i=1}^{15}((-1)^ii)^2\\ =\left(\sum_{i=1}^{15}(-1)^ii\right)^2-\sum_{i=1}^{15}i^2\\ =8^2-1240.$$