Write the 4th term from the end in the expansion of

To find: $4^{\text {th }}$ term from the end in the expansion of $\left(\frac{3}{x^{2}}-\frac{x^{3}}{6}\right)^{7}$

Formula Used:

A general term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by,

$T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where

${ }^{\mathrm{n}} \mathrm{C}_{r}=\frac{n !}{r !(n-r) !}$

Total number of terms in the expansion is 8

Thus, the $4^{\text {th }}$ term of the expansion is $T_{5}$ and is given by,

$T_{5}={ }^{7} C_{5} \times\left(\frac{3}{x^{2}}\right)^{3} \times\left(\frac{-x^{3}}{6}\right)^{4}$

$T_{5}=\frac{7 \times 6 \times 5 !}{2 \times 5 !} \times \frac{3 \times 3 \times 3}{6 \times 6 \times 6 \times 6} \times x^{-18}$

$T_{5}=\frac{7 \times 6 \times 5 !}{2 \times 5 !} \times \frac{3 \times 3 \times 3}{6 \times 6 \times 6 \times 6} \times x^{-18}$

$T_{5}=\frac{7}{16} \mathrm{x}^{-18}$

Thus, a $4^{\text {th }}$ term from the end in the expansion of $\left(\frac{3}{x^{2}}-\frac{x^{3}}{6}\right)^{7}$ is $T_{5}=\frac{7}{16} x^{-18}$