Find the \( 15 t \) terms of the ff \( (1-1 / 3 x)^{-3} \)

To find the \(15t\) term of the series expansion for \((1 - \frac{1}{3}x)^{-3}\), we can use the binomial series expansion, which states that: \[ (1-u)^{-n} = \sum_{k=0}^{\infty} \binom{n+k-1}{k} u^k \] In our case, \(n = 3\) and \(u = \frac{1}{3}x\). Hence, we can express the series as: \[ (1 - \frac{1}{3}x)^{-3} = \sum_{k=0}^{\infty} \binom{3+k-1}{k} \left(\frac{1}{3}x\right)^k = \sum_{k=0}^{\infty} \binom{k+2}{2} \left(\frac{1}{3}\right)^k x^k \] The \(k\)-th term in this expansion is given by: \[ \binom{k+2}{2} \frac{1}{3^k} x^k \] To find the \(15t\) term, we want the coefficient of \(x^{15}\) in this series expansion, which corresponds to \(k=15\). So we can compute: \[ \binom{15+2}{2} \frac{1}{3^{15}} = \binom{17}{2} \frac{1}{3^{15}} = \frac{17 \times 16}{2} \frac{1}{3^{15}} = 136 \frac{1}{3^{15}} \] Thus, the \(15t\) term of the series \((1 - \frac{1}{3} x)^{-3}\) is: \[ \frac{136}{3^{15}} x^{15} \]