Let the coefficients of the middle terms in the expansion of (1√6+ x)^4, (1-3 x)^2 and (1-…

Question

Let the coefficients of the middle terms in the expansion of $\left(\dfrac{1}{\sqrt6}+\beta x\right)^4$, $(1-3\beta x)^2$ and $\left(1-\dfrac{\beta}{2}x\right)^6$, $\beta>0$, respectively form the first three terms of an A.P. If $d$ is the common difference of this A.P., then $50-\dfrac{2d}{\beta^2}$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 57

Accepted Answer

Middle-term coefficients: $\binom42\dfrac1{6}\beta^2$, $\binom21(-3\beta)$, $\binom63\left(-\dfrac{\beta}{2}\right)^3$ form an A.P., leading to

$$5\beta^2-2\beta-24=0\quad\Rightarrow\quad(5\beta-12)(\beta+2)=0\quad\Rightarrow\quad\beta=\frac{12}{5}\ (\beta>0)$$

$$d=-6\beta-\beta^2\quad\Rightarrow\quad50-\frac{2d}{\beta^2}=50+\frac{12}{\beta}+2=57$$

Solutions · 1