(a) To expand \(\left( x + \frac{2}{x} \right)^5\), use the binomial theorem:
\(\sum_{k=0}^{5} \binom{5}{k} x^{5-k} \left( \frac{2}{x} \right)^k\).
Calculate each term:
\(\binom{5}{0} x^5 = x^5\)
\(\binom{5}{1} x^4 \cdot \frac{2}{x} = 10x^3\)
\(\binom{5}{2} x^3 \cdot \left( \frac{2}{x} \right)^2 = 40x\)
\(\binom{5}{3} x^2 \cdot \left( \frac{2}{x} \right)^3 = \frac{80}{x}\)
\(\binom{5}{4} x \cdot \left( \frac{2}{x} \right)^4 = \frac{80}{x^3}\)
\(\binom{5}{5} \left( \frac{2}{x} \right)^5 = \frac{32}{x^5}\)
Thus, the expansion is \(x^5 + 10x^3 + 40x + \frac{80}{x} + \frac{80}{x^3} + \frac{32}{x^5}\).
(b) Consider the expansion \((a + bx^2) \left( x + \frac{2}{x} \right)^5\).
The coefficient of \(x\) is zero:
\(40a + (\text{coefficient of } x^{-1})b = 0\)
The coefficient of \(\frac{1}{x}\) is 80:
\((\text{coefficient of } x^{-1})a + (\text{coefficient of } x^{-3})b = 80\)
From part (a), the coefficient of \(x^{-1}\) is 80 and \(x^{-3}\) is 80.
Thus, the equations are:
\(40a + 80b = 0\)
\(80a + 80b = 80\)
Solving these equations:
From \(40a + 80b = 0\), we get \(a = -2b\).
Substitute into \(80a + 80b = 80\):
\(80(-2b) + 80b = 80\)
\(-160b + 80b = 80\)
\(-80b = 80\)
\(b = -1\)
Substitute \(b = -1\) into \(a = -2b\):
\(a = 2\)
Thus, \(a = 2\) and \(b = -1\).
