Discussion
In the binomial expansion of \((2x + \alpha)^{8}\) , the co-efficients of \(x^{2}\) and \(x^{3}\) are equal. Then the value of \(\alpha\) is equal to
Answer: C
\(\alpha = \) 4.
The general term of \((2x + \alpha)^{8}\) is \({}^{8}C_{r}(2x)^{r}\alpha^{8 - r}\). So
\[ \begin{aligned} \text{coeff of } x^{2} &= {}^{8}C_{2}\cdot 4 \cdot \alpha^{6} = 112\alpha^{6} \\ \text{coeff of } x^{3} &= {}^{8}C_{3}\cdot 8 \cdot \alpha^{5} = 448\alpha^{5} \end{aligned} \]
Setting them equal (with \(\alpha \neq 0\)) gives \(112\alpha = 448\), so \(\alpha = 4\). Forgetting the powers of 2 from \((2x)^{r}\) gives \(28\alpha = 56\), \(\alpha = 2\); inverting the ratio gives 1/4.
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