(i) To find the first three terms in the expansion of \((2 + ax)^5\), we use the binomial theorem:
\((2 + ax)^5 = \sum_{k=0}^{5} \binom{5}{k} (2)^{5-k} (ax)^k\).
The first three terms are:
\(\binom{5}{0} (2)^5 (ax)^0 = 32\),
\(\binom{5}{1} (2)^4 (ax)^1 = 80ax\),
\(\binom{5}{2} (2)^3 (ax)^2 = 80a^2x^2\).
Thus, the first three terms are \(32 + 80ax + 80a^2x^2\).
(ii) We need to find the coefficient of \(x^2\) in the expansion of \((1 + 2x)(2 + ax)^5\).
First, expand \((2 + ax)^5\) to get the terms involving \(x^2\):
\(32 + 80ax + 80a^2x^2\).
Now, multiply by \((1 + 2x)\):
\((32 + 80ax + 80a^2x^2)(1 + 2x)\).
The terms contributing to \(x^2\) are:
\(32 \cdot 2x = 64x\),
\(80ax \cdot 1 = 80ax\),
\(80a^2x^2 \cdot 1 = 80a^2x^2\).
The coefficient of \(x^2\) is \(80a^2 + 160a\).
Set this equal to 240:
\(80a^2 + 160a = 240\).
Divide by 80:
\(a^2 + 2a = 3\).
Rearrange to form a quadratic equation:
\(a^2 + 2a - 3 = 0\).
Factorize:
\((a - 1)(a + 3) = 0\).
Thus, \(a = 1\) or \(a = -3\).
