Discussion
If the coefficient of \(y^{3}\) in the binomial expansion of \(\left(2\alpha - \frac{y}{2}\right)^{8}\) is -7 , then the value of \(\alpha\) is equal to
Answer: A
The value is \(\alpha = \frac{1}{2}\).
The general term of \((2\alpha - \frac{y}{2})^{8}\) is \(\binom{8}{r}(2\alpha)^{8-r}\left(-\frac{y}{2}\right)^{r}\). For \(y^{3}\), r = 3.
\[ \begin{aligned} T_4 &= \tbinom{8}{3}(2\alpha)^{5}\left(-\tfrac{1}{2}\right)^{3}y^{3} \\ &= 56 \times 32\alpha^{5} \times \left(-\tfrac{1}{8}\right)y^{3} \\ &= -224\alpha^{5}y^{3} \end{aligned} \]
Setting \(-224\alpha^{5} = -7\) gives \(\alpha^{5} = \frac{1}{32}\), so \(\alpha = \frac{1}{2}\). Dropping the minus sign from \(\left(-\tfrac{1}{2}\right)^{3}\) would make \(\alpha\) negative, and the whole-number options make the coefficient far too large (\(\alpha = 2\) gives -7168).
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