(i) To find the term independent of \(x\) in the expansion of \(\left( \frac{2}{x} - 3x \right)^6\), we use the binomial theorem:
\(\binom{6}{3} \left( \frac{2}{x} \right)^3 (-3x)^3\)
Calculating each part:
\(\binom{6}{3} = 20\)
\(\left( \frac{2}{x} \right)^3 = \frac{8}{x^3}\)
\((-3x)^3 = -27x^3\)
Multiplying these gives:
\(20 \times \frac{8}{x^3} \times (-27x^3) = -4320\)
Thus, the term independent of \(x\) is \(-4320\).
(ii) To find the value of \(a\) for which there is no term independent of \(x\) in the expansion of \(\left( 1 + ax^2 \right) \left( \frac{2}{x} - 3x \right)^6\), we consider the critical term:
\(\binom{6}{2} \left( \frac{2}{x} \right)^4 (-3x)^2\)
Calculating each part:
\(\binom{6}{2} = 15\)
\(\left( \frac{2}{x} \right)^4 = \frac{16}{x^4}\)
\((-3x)^2 = 9x^2\)
Multiplying these gives:
\(15 \times \frac{16}{x^4} \times 9x^2 = \frac{2160}{x^2}\)
For no term independent of \(x\), the coefficient of \(x^0\) must be zero:
\(15a \times 16 \times 9 - 4320 = 0\)
Solving for \(a\):
\(2160a = 4320\)
\(a = 2\)
