The coefficient of $x^{18}$ in the expansion of $\left(x^{4}-\frac{1}{x^{3}}\right)^{15}$ is __________.
Answer (integer)
5005
Solution
$\begin{aligned} T_{r+1}= & { }^{15} C_r\left(x^4\right)^{15-r}\left(-\frac{1}{x^3}\right)^r={ }^{15} C_r(-1)^r x^{60-4 r-3 r} \\\\ = & { }^{15} C_r(-1)^r x^{60-7 r}\end{aligned}$
<br/><br/>$\begin{aligned} \therefore 60-7 r =18 \\\\ \Rightarrow 7 r =42 \\\\ \Rightarrow r =6\end{aligned}$
<br/><br/>$\therefore$ The coefficient of $x^{18}$
<br/><br/>$$
={ }^{15} C_6(-1)^6=\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{6 \times 5 \times 4 \times 3 \times 2 \times 1}=5005
$$
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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