Question

The coefficient of \(x^{18}\) in the expansion of \(\left(x^4 - \frac{1}{x^3}\right)^{15}\) is ________

Answer 1

Correct Answer: 6

The answer is \(6\)
\(T_{r+1}=^{15}C_{r}(x^{4})^{15-r}(-\frac{1}{x^{3}})^{r}\)
\(T_{r+1}=^{15}C_{r}(-1)^{r}x^{60-7r}\)
\(60-7r=18\Rightarrow r=6\)
\(T_{7}=^{15}C_{6}(-1)^{6}x^{18}\)
\(T_{7}=^{15}C_{6}x^{18}\)
So, the Coefficient of x¹⁸ is ¹⁵C₆

Answer 2

The general term in the binomial expansion of \((a+b)^n\) is given by \(T_{r+1} = \binom{n}{r} a^{n-r} b^r\). In the expansion of \(\left(x^4 - \frac{1}{x^3}\right)^{15}\), the \((r+1)\)th term is given by: \[ T_{r+1} = \binom{15}{r} (x^4)^{15-r} \left(-\frac{1}{x^3}\right)^r = \binom{15}{r} x^{60-4r} (-1)^r x^{-3r} = \binom{15}{r} (-1)^r x^{60-7r}. \] We are looking for the coefficient of \(x^{18}\). Therefore, we need to find \(r\) such that \(60 - 7r = 18\). \[ 60 - 7r = 18 \Rightarrow 7r = 42 \Rightarrow r = 6. \] So, the term with \(x^{18}\) is \[ T_{6+1} = T_7 = \binom{15}{6}(-1)^6 x^{18} = \binom{15}{6} x^{18}. \] The coefficient of \(x^{18}\) is \(\binom{15}{6} = \frac{15!}{6!9!} = 5005\).