<s>[INST] Bi-nominal expansion of 3 terms Find the coefficient of $x^{17}$ in the expansion of $(3x^7 + 2x^5 -1)^{20}$
I'm stuck in handling this question as I do know how to solve it when it has 2 terms.
But now it has 3.
I have no idea where to begin...
 [/INST] We can apply the binomial theorem twice in order to find the coeffcient. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of an expression.

We obtain
  \begin{align*}
\color{blue}{[x^{17}]}&\color{blue}{(3x^7+2x^5-1)^{20}}\\
&=[x^{17}]\sum_{k=0}^{20}\binom{20}{k}(3x^7)^k(2x^5-1)^{20-k}\tag{1}\\
&=\binom{20}{0}3^0[x^{17}](2x^5-1)^{20}+\binom{20}{1}3^1[x^{10}](2x^5-1)^{19}\\
&\qquad +\binom{20}{2}3^2[x^3](2x^5-1)^{18}\tag{2}\\
&=3\binom{20}{1}[x^{10}]\sum_{j=0}^{19}\binom{19}{j}(2x^5)^j(-1)^{19-j}\tag{3}\\
&=-3\binom{20}{1}\binom{19}{2}2^2\tag{4}\\
&\,\,\color{blue}{=-41\,040}
\end{align*}

Comment:


*

*In (1) we apply the binomial theorem once to $(3x^7+(2x^5-1))^{20}$.

*In (2) we observe that the term $(3x^7)^k$ contributes to $[x^{17}]$ only when $k=0,1,2$. We also apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.

*In (3) we note that expanding $(2x^5-1)^n$ gives only exponents of $x$ which are a multiple of $5$. We can therefore stick to the middle summand with $[x^{10}]$ and we apply the binomial theorem again.

*In (4) we select the coefficient of $x^j$ accordingly.
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