Question

Find the third term in the expansion of (2x + 1/(3x ^ 2)) ^ 9​

Answer 1

Step-by-step explanation:

$\text{Here \( \tt a=9 x^{2}, b=\dfrac{-y^{3}}{6}, n=4 \)}$

$\text{ For \( \tt 3^{\text {rd }} \) term, \( \tt r=2 \)}$

$\text{We have, \( \tt t_{r+1}={}^{n}C_{r} a^{n - r} \cdot b^{r} \)}$

$\pmb{ \begin{aligned} \tt \therefore \quad t_{3} & \tt={ }^{4} C_{2}\left(9 x^{2}\right)^{4-2}\left(\frac{-y^{3}}{6}\right)^{2} \\ \\ & \tt=\frac{4 !}{2 ! 2 !}\left(81 x^{4}\right) \cdot \frac{y^{6}}{36} \\ \\ & \tt=\frac{4 \times 3}{2 \times 1} \times 81 x^{4} \times \frac{y^{6}}{36}-\frac{27}{2} x^{4} y^{6} \end{aligned} }$

$\tt 3^{\text {rd }}$ term in the expansion of $\tt\left(9 x^{2}-\dfrac{y^{3}}{6}\right)^{4}$ is $\tt \dfrac{27}{2} x^{4} y^{6}.$