If the coefficients of x^-2 and x^-4 in the expansion of (x^1/3+12x^1/3)^18, (x0), are m …

Question

If the coefficients of $x^{-2}$ and $x^{-4}$ in the expansion of $\left(x^{1/3}+\dfrac{1}{2x^{1/3}}\right)^{18}$, ($x>0$), are $m$ and $n$ respectively, then $\dfrac mn$ is equal to:

Accepted Answer

$$T_{r+1}=\binom{18}{r}\left(x^{1/3}\right)^{18-r}\left(\frac{1}{2x^{1/3}}\right)^r=\binom{18}{r}\left(\frac12\right)^r\,x^{6-\frac{2r}{3}}$$

For $x^{-2}$: $6-\dfrac{2r}3=-2\Rightarrow r=12$: $m=\binom{18}{12}\left(\dfrac12\right)^{12}$

For $x^{-4}$: $6-\dfrac{2r}3=-4\Rightarrow r=15$: $n=\binom{18}{15}\left(\dfrac12\right)^{15}$

$$\frac mn=\frac{\binom{18}{12}}{\binom{18}{15}}\times2^3=\frac{18!/(12!6!)}{18!/(15!3!)}\times8=\frac{15!3!}{12!6!}\times8=\frac{13\times14\times15}{4\times5\times6}\times8=\frac{2730}{120}\times8=22.75\times8=182$$

Solutions · 1