The coefficient of x^18 in the expansion of (x^4-1x^3)^15 is __________.

Question

The coefficient of ${x}^{18}$ in the expansion of ${({x}^{4}-\frac{1}{{x}^{3}})}^{15}$ is \_\_\_\_\_\_\_\_\_\_.

Answer: 5005

Accepted Answer

$\begin{gathered} {T}_{r+1}= {}^{15}{C}_{r}{({x}^{4})}^{15-r}{(-\frac{1}{{x}^{3}})}^{r}={}^{15}{C}_{r}(-1{)}^{r}{x}^{60-4r-3r} \ \ = {}^{15}{C}_{r}(-1{)}^{r}{x}^{60-7r} \end{gathered}$

$\begin{gathered} 	herefore 60-7r=18 \ \Rightarrow 7r=42 \ \Rightarrow r=6 \end{gathered}$

$	herefore $ The coefficient of ${x}^{18}$

$={}^{15}{C}_{6}(-1{)}^{6}=\frac{15	imes 14	imes 13	imes 12	imes 11	imes 10}{6	imes 5	imes 4	imes 3	imes 2	imes 1}=5005$

Solutions · 1