Find the coefficient of $x^{70}$ in $(x^1-1)(x^2-2)(x^3-3)\cdots (x^{12}-12)$.
I tried to solve this problem using theory of equation the coefficient of $x^{70}$ will be the sum of products taking two at a time. But this very very exhaustive I want to know some another method as it will be proficient in higher powers.
$$1+2+\ldots+12=\frac{12\times 13}{2}=78$$
$78-8=70$ gives contribution of $-8$
$78-7-1=70$ gives $(-7)(-1)$
$78-6-2=70$ gives $(-6)(-2)$
$78-5-3=70$ gives $(-5)(-3)$
$78-5-2-1=70$ gives $(-5)(-2)(-1)$
$78-4-3-1=70$ gives $(-4)(-3)(-1)$
Can you proceed?
Since $1+2+3+\ldots+12=78$, the term $x^{70}$ must arise from taking $x^k$ from $(x^k-k)$ for almost every $k\in\{1,2,\ldots,12\}$, except for some $j_1,j_2,\ldots,j_r\in\{1,2,\ldots,k\}$ such that $j_1<j_2<\ldots<j_r$ and $j_1+j_2+\ldots+j_r=8$. There are very few such tuples $(j_1,j_2,\ldots,j_r)$:
for $r=1$, $j_1=8$;
for $r=2$, $(j_1,j_2)=(1,7),(2,6),(3,5)$;
for $r=3$, $(j_1,j_2,j_3)=(1,2,5),(1,3,4)$.
Therefore, the coefficient of $x^{70}$ is $$(-1)^1\cdot 8+(-1)^2\cdot (1\cdot 7+2\cdot 6+3\cdot 5)+(-1)^3\cdot(1\cdot 2\cdot 5+1\cdot 3\cdot 4)=4\,.$$
For any formal Laurent series $f(z) = \sum_{k=-\infty}^\infty a_k z^k$ in any variable $z$, let $[z^k]f(z)$ be a short hand for the coefficient $a_k$.
Let $y = x^{-1}$, we can simplify the coefficient at hand as
$$\begin{align}\mathcal{C} \stackrel{def}{=} [x^{70}]\prod_{k=1}^{12}(x^k-k) &= [x^{70}]x^{78} \prod_{k=1}^{12}(1-kx^{-k}) \\ &= [y^8]\prod_{k=1}^{12}(1-ky^k) = [y^8]\prod_{k=1}^8(1-ky^k)\end{align}$$
We can split last product into two pieces
$$\begin{align}\prod_{k=1}^3 (1-ky^k) &= (1-y)(1-2y^2)(1-3y^3) = (1-y-2y^2+2y^3)(1-3y^3)\\ &= 1-y-2y^2-y^3 + 3y^4 +6y^5 - 6y^6\\ \prod_{k=4}^8 (1-ky^k) &= 1 - 4y^4 - 5y^5 - 6y^6 - 7y^7 - 8y^8 + O(y^9) \end{align}$$ Throwing away terms which doesn't contribute to $[y^8]$, we obtain $$\begin{align}\mathcal{C} = &\; [y^8]\big((1-y-2y^2-y^3+3y^4)(1-4y^4-5y^5-6y^6-7y^7-8y^8)\big)\\ = &\; (1)(-8) + (-1)(-7) + (-2)(-6)+(-1)(-5)+(3)(-4)\\ = &\; -8 + 7 + 12 + 5 - 12\\ = &\; 4\end{align}$$
If we multiply the powers of $x$, we get $x^{1+...+12}=x^{78}.$ We are looking for the ways we can multiply the powers of $x$ together in order to get a result of $x^{70}$. Note that we must remove 8 from 1+...+12. Some ways to do this are 1+...+7+9+...+12, 1+3+...+5+7+...+12, 3+4+6+...+12. These will contribute -8, (-2)(-6),(-1)(-2)(-5) (additively) to the coefficient you are looking for. Now we need to find the rest.
What we are looking for are called partitions of 8 (without repeated numbers).
$\prod\limits_{i=1}^n (a_i + b_i) = \sum\limits_{s\in \{0,1\}^n}[ \prod\limits_{i=1}^n \begin{cases} a_i & \text{if }s_i =0\\b_i &\text{if }s_i = 1\end{cases}]$
(Thats confusingly written by I hope it's clear. $\prod\limits_{i=1}^n (a_i + b_i)$ will be the sum of the products when you chose either $a_i$ or $b_i$ for any combination.)
So $\prod\limits_{i=1}^{12} (x^i - 1)= \sum\limits_{K\subset \{1,2,...,12\}} (\prod_{i\in K}x^i\prod\limits_{j\not \in K} (-j))=$
$=\sum\limits_{K\subset \{1,2,...,12\}}( x^{\sum\limits_{i\in K} i}\prod\limits_{j\not \in K} (-j))=$
$\sum\limits_{k=0...78} x^k(\sum\limits_{K\subset\{1,...,12\}| \sum\limits_{j\in K} j = k}\prod\limits_{m\not \in K}(-m))$
Which is to say for all the ways to add $1,...., 12$ to get $70$ the coefficienct will be the sum of the products of the negatives of all the other numbers.
And as the ways to add up to $70$ is precisely the ways omit all the numbers that add to $8$, the coefficient will be the sum of the products of the negatives of all the ways to add numbers to get $8$
i.e. $(-8) + (-1*-7) + (-2*-6) + (-3*-5) + (-1*-2*-5)+(-1*-3*-4)=$
$-8 + 7 + 12 + 15 -10 - 12 = 4$.
And... I probably made a horrible mistake. Didn't I?
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in other words: $(x-1)(x^2 - 2).....(x^{12} - 12) =x^1x^2...x^{12}+ .... +x^{1}x^2...x^{7}x^9...x^{12}(-8) + x^2x^3...x^6,x^7..x^{12}(-1)(-7) + ..... +(-1)(-2)...(-12)=x^{78} + ...... +x^{70}(-8 + (-1)(-7)+.......)+.....+12!$
