To find the coefficient of \(\frac{1}{x^2}\) in the expansion of \(\left( 3x + \frac{2}{3x^2} \right)^7\), we use the binomial theorem:
\(\left( 3x + \frac{2}{3x^2} \right)^7 = \sum_{k=0}^{7} \binom{7}{k} (3x)^{7-k} \left( \frac{2}{3x^2} \right)^k\)
We need the term where the power of \(x\) is \(-2\):
\((3x)^{7-k} \left( \frac{2}{3x^2} \right)^k = 3^{7-k} x^{7-k} \cdot \frac{2^k}{3^k x^{2k}} = \frac{3^{7-k} \cdot 2^k}{3^k} x^{7-k-2k}\)
Set the power of \(x\) to \(-2\):
\(7 - 3k = -2\)
\(3k = 9\)
\(k = 3\)
Substitute \(k = 3\) into the expression:
\(\binom{7}{3} (3x)^{4} \left( \frac{2}{3x^2} \right)^3\)
\(= \binom{7}{3} \cdot 3^4 \cdot \frac{2^3}{3^3} \cdot x^{4-6}\)
\(= 35 \cdot 81 \cdot \frac{8}{27} \cdot x^{-2}\)
\(= 35 \cdot 3 \cdot 8 \cdot x^{-2}\)
\(= 840 \cdot x^{-2}\)
Thus, the coefficient of \(\frac{1}{x^2}\) is 840.
