Question

Find the 5th term from the end in the expansion of ${(3x-\frac{1}{x^2})}^{10}$.

Answer 1

${(3x-\frac{1}{x^2})}^{10}$

As we know that generalt term of expnsion ${(a+b)}^n$ is-

$T_{r+1}={}_n{C}_r{a}^{n-r}{b}^n$

In the given expansion-

${(3x-\frac{1}{x^2})}^{10}$

$n=10$

$a=3x$

$b=-\frac{1}{x^2}$

Therefore, general term of the given epansion will be-

$T_{r+1}={}_{10}{C}_r{\left(3x\right)}^{10-r}{\left(\frac{1}{x^2}\right)}^r$

$\Rightarrow T_{r+1}={}_{10}{C}_r\left(3^{10-r}\right)\left(x^{10-r}\right)\left(\frac{1}{x^{2r}}\right)$

$\Rightarrow T_{r+1}={}_{10}{C}_r\left(3^{10-r}\right)\left(x^{10-3r}\right).....\left(1\right)$

As we know that,

$p^{th}$ term form end $={(n-p+2)}^{th}$ term from starting.

$\therefore 5^{th}$ term form end $={(10-5+2)}^{th}=7^{th}$ term from starting.

$\therefore T_7=T_{6+1}$

$\Rightarrow r=6$

Substituting the value of $r$ in ${eq}^n\left(1\right)$, we have

$T_{6+1}={}_{10}{C}_6\left(3^{10-6}\right)\left(x^{10-(3\times 6)}\right)$

$\Rightarrow T_7={}_{10}{C}_6\left(3^4\right)\left(x^{-8}\right)$

Hence the $5^{th}$ term from the end in the given expansion will be $7^{th}$ term from the starting, i.e., ${}_{10}{C}_6\left(3^4\right)\left(x^{-8}\right)$.