Question

Find the 4th term from the end in the expansion of ${(\frac{3}{x^2}-\frac{x^3}{6})}^7$.

Answer 1

We know that, r th term from the end of $(x+y{)}^n$ is $T_{n+2-r}$

Here, n=7 and r=4

So, 4 th term from the end is $T_{7+2-4}=T_5$

$\therefore$ required term, $T_5=7_{C_4}{\left(\frac{3}{x^2}\right)}^3(-1{)}^4{\left(\frac{x^3}{6}\right)}^6=\frac{7!}{3!4!}\times \frac{27}{x^6}\times \frac{x^{12}}{6^4}=\frac{35}{48}{x}^6$