How to find Coefficient in product of binomials?

Question

How can i find coefficient of $x^{51}$ in expansion of $$(x-1)(x^2-2)(x^3-3)\ldots (x^{10}-10)?$$ I tried all methods and formulae but couldn't find the right answer.

Answer 1

Note that $1+2+\cdots+10=55$.

When you multiply out this polynomial, you will get a sum of terms; each term is the result of choosing either one term or the other from each binomial, and multiplying all of them together.

So, you get an $x^{51}$ by choosing $x^{m}$ terms whose exponents add up to $51$, and using the constant terms for the others.

Now, because $1+2+\cdots+10=55$, you can equivalently consider this as choosing to use constant terms in place of $x^m$'s where the $m$'s add up to $4$. How many different ways can you choose a subset of $\{1,2,3,\ldots,10\}$ that add up to $4$? (Not very many!)

Answer 2

To expand on the answer by Nicholas R. Peterson, you can perform the following operations: put $y=x^{-1}$, multiply each factor $x^d-d$ by$~y^d$ (giving $1-dy^d$) and now the term in the expansion of the product of degree $55-k$ in $x$ has become a term of degree $k$ in$~y$. So concretely you are asking for the coefficient of $y^4$ in the product $\prod_{d\geq1}(1-dy^d)$; this is easy to compute.

I have made the product infinite; this does not change the coefficients in degrees${}\leq10$ (so it still gives an answer to your question), because the infinite product is mathematically more interesting. Nevertheless, it does not seem that interesting; OEIS knows nothing about its sequence of coefficients $1,-1,-2,-1,-1,5,1,13,4,2,-8,-61,-31,13,-156,21,11,223,\ldots$