If the coefficients of x^7 in (ax^2+12bx)^11 and x^-7 in (ax-13bx^2)^11 are equal, then:

Question

If the coefficients of $x^7$ in $\left(ax^2+\dfrac{1}{2bx}\right)^{11}$ and $x^{-7}$ in $\left(ax-\dfrac{1}{3bx^2}\right)^{11}$ are equal, then:

Accepted Answer

For $\left(ax^2+\dfrac{1}{2bx}\right)^{11}$: power $=2(11-r)-r=22-3r=7\Rightarrow r=5$. Coefficient $=\binom{11}5a^6\left(\dfrac{1}{2b}\right)^5$.

For $\left(ax-\dfrac{1}{3bx^2}\right)^{11}$: power $=(11-r)-2r=11-3r=-7\Rightarrow r=6$. Coefficient $=\binom{11}6a^5\left(-\dfrac{1}{3b}\right)^6$.

Equating (using $\binom{11}5=\binom{11}6$): $a^6\cdot\dfrac{1}{2^5b^5}=a^5\cdot\dfrac{1}{3^6b^6}\Rightarrow ab=\dfrac{2^5}{3^6}\Rightarrow729ab=32$

Solutions · 1