Mathematics - Binomial Theorem Question with Solution | TestHub

For ${\left(3x^3-2x^2+\frac{5}{x^5}\right)}^{10}$the general term is given by

$T_{r+1}=\frac{10!}{r_1!r_2!r_3!}{\left(3\right)}^{r_1}{\left(-2\right)}^{r_2}{\left(5\right)}^{r_3}{\left(x\right)}^{3r_1+2r_2-5r_3}$

For term independent of $x$ the exponent $3r_1+2r_2-5r_3=0 \dots \left(1\right)$

Also we know $r_1+r_2+r_3=10 \cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get 

$r_1+2\left(10-r_3\right)-5r_3=0$

i.e. $r_1+20=7r_3$

So $r_1,r_2,r_3=1,6,3$

Hence the constant term $=\frac{10!}{1!6!3!}{\left(3\right)}^1{\left(-2\right)}^6{\left(5\right)}^3$

$=2^9·3^2·5^4·7^1$

$\Rightarrow k=9$