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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.

Since $1+2+3+ldots+12=78$, the term $x^{70}$ must arise from taking $x^k$ from $(x^k-k)$ for almost every $kin{1,2,ldots,12}$, except for some $j_1,j_2,ldots,j_rin{1,2,ldots,k}$ 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
$$(-1)^1cdot 8+(-1)^2cdot (1cdot 7+2cdot 6+3cdot 5)+(-1)^3cdot(1cdot 2cdot 5+1cdot 3cdot 4)=4,.$$

$$1+2+ldots+12=frac{12times 13}{2}=78$$

• $78-8=70$ 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?

Check with the Wolfram Alpha output

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
$$begin{align}mathcal{C} = &; [y^8]big((1-y-2y^2-y^3+3y^4)(1-4y^4-5y^5-6y^6-7y^7-8y^8)big)\
= &; (1)(-8) + (-1)(-7) + (-2)(-6)+(-1)(-5)+(3)(-4)\
= &; -8 + 7 + 12 + 5 - 12\
= &; 4end{align}$$

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).

$prodlimits_{i=1}^n (a_i + b_i) = sumlimits_{sin {0,1}^n}[ prodlimits_{i=1}^n begin{cases} a_i & text{if }s_i =0\b_i &text{if }s_i = 1end{cases}]$

(Thats confusingly written by I hope it's clear. $prodlimits_{i=1}^n (a_i + b_i)$ will be the sum of the products when you chose either $a_i$ or $b_i$ for any combination.)

So $prodlimits_{i=1}^{12} (x^i - 1)= sumlimits_{Ksubset {1,2,...,12}} (prod_{iin K}x^iprodlimits_{jnot in K} (-j))=$

$=sumlimits_{Ksubset {1,2,...,12}}( x^{sumlimits_{iin K} i}prodlimits_{jnot in K} (-j))=$

$sumlimits_{k=0...78} x^k(sumlimits_{Ksubset{1,...,12}| sumlimits_{jin K} j = k}prodlimits_{mnot in K}(-m))$

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: $(x-1)(x^2 - 2).....(x^{12} - 12) =x^1x^2...x^{12}+ .... +x^{1}x^2...x^{7}x^9...x^{12}(-8) + x^2x^3...x^6,x^7..x^{12}(-1)(-7) + ..... +(-1)(-2)...(-12)=x^{78} + ...... +x^{70}(-8 + (-1)(-7)+.......)+.....+12!$