If the coefficient of x^7 in (ax-1bx^2)^13 and the coefficient of x^-5 in (ax+1bx^2)^13 ar…

Question

If the coefficient of $x^7$ in $\left(ax-\dfrac{1}{bx^2}\right)^{13}$ and the coefficient of $x^{-5}$ in $\left(ax+\dfrac{1}{bx^2}\right)^{13}$ are equal, then $a^4b^4$ is equal to:

Accepted Answer

For $\left(ax-\dfrac{1}{bx^2}ight)^{13}$: $T_{r+1}=\binom{13}{r}a^{13-r}(-1)^rb^{-r}x^{13-3r}$. For $x^7$: $r=2$, coefficient $=\binom{13}{2}a^{11}b^{-2}$.

For $\left(ax+\dfrac{1}{bx^2}ight)^{13}$: $T_{r+1}=\binom{13}{r}a^{13-r}b^{-r}x^{13-3r}$. For $x^{-5}$: $r=6$, coefficient $=\binom{13}{6}a^7b^{-6}$.

Equating: $\binom{13}{2}a^{11}b^{-2}=\binom{13}{6}a^7b^{-6}\Rightarrow a^4b^4=\dfrac{\binom{13}{6}}{\binom{13}{2}}=\dfrac{13	imes12	imes11	imes10	imes9	imes8}{13	imes12	imes6	imes5	imes4	imes3}=22$

Solutions · 1