Finding coefficient in the expansion

Question

Given :

$(x -1)(x^2 -2 )(x^3 - 3)......(x^{11} -11)$

Find the coefficient of $x^{60}$ in the expansion

My attempt: I tried to expand it term by term but it became huge and weird. Please help me to solve this. Thanks in advance

Answer 1

The highest power that can be achieved is $x^{66}$ and we want $x^{60}$. So we find the brackets from which we select the constant terms such that the sum of powers of $x$ of those brackets is $6$. Hence we select the $x$'s from all other brackets except for them.

The brackets having the powers of x summing to 6 are $-6$ ;$(-1)$ and $(-5)$; $(-2)$ and $(-4)$; $(-1),(-2)$ and $(-4)$

so the needed coefficient is

$$(-6)+(-1)(-5)+(-2)(-4)+(-1)(-2)(-3)=1$$

Answer 2

Coefficient of $x^{60}$ in $$x^{1+2+3+\cdots +11}\bigg[\bigg(1-\frac{1}{x}\bigg)\bigg(1-\frac{2}{x^2}\bigg)\bigg(1-\frac{3}{x^3}\bigg)\cdots \bigg(1-\frac{11}{x^{11}}\bigg)\bigg]$$

So coefficient of $x^{-6}$ in $$\bigg[\bigg(1-\frac{1}{x}\bigg)\bigg(1-\frac{2}{x^2}\bigg)\bigg(1-\frac{3}{x^3}\bigg)\cdots \bigg(1-\frac{11}{x^{11}}\bigg)\bigg]$$

To get $x^{-6}$ we should take powers of from each bracket except for the few ones that add up to . These are the only ways to get $6$ with different positive integers

$6=6,6=1+5,6=2+4,6=1+2+3$ and so the wanted coefficients is

$$(-6)+(-1)(-5)+(-2)(-4)+(-1)(-2)(-3)=1$$