If the coefficient of x^9 in ( x^3+1 x)^11 and the coefficient of x^-9 in ( x-1 x^3)^11 ar…

Question

If the coefficient of $x^9$ in $\left(\alpha x^3+\dfrac{1}{\beta x}\right)^{11}$ and the coefficient of $x^{-9}$ in $\left(\alpha x-\dfrac{1}{\beta x^3}\right)^{11}$ are equal, then $(\alpha\beta)^2$ is equal to \_\_\_\_\_\_\_.

Answer: 1

Accepted Answer

Coefficient of $x^9$ in $\left(\alpha x^3+\dfrac{1}{\beta x}\right)^{11}$: general term $\binom{11}{r}\alpha^{11-r}\beta^{-r}x^{3(11-r)-r}=x^{33-4r}$. For $33-4r=9\Rightarrow r=6$: coefficient $=\binom{11}{6}\alpha^5\beta^{-6}$.

Coefficient of $x^{-9}$ in $\left(\alpha x-\dfrac{1}{\beta x^3}\right)^{11}$: general term $\binom{11}{r}\alpha^{11-r}(-1)^r\beta^{-r}x^{(11-r)-3r}=x^{11-4r}$. For $11-4r=-9\Rightarrow r=5$: coefficient $=-\binom{11}{5}\alpha^6\beta^{-5}$.

Setting them equal: $\binom{11}{6}\alpha^5\beta^{-6}=-\binom{11}{5}\alpha^6\beta^{-5}$. Since $\binom{11}{6}=\binom{11}{5}$: $\dfrac{1}{\beta}=-\alpha\Rightarrow\alpha\beta=-1\Rightarrow(\alpha\beta)^2=1$

Solutions · 1