Mathematics - Binomial Theorem Question with Solution | TestHub

We know that general term in the expansion of ${\left(x-\frac{3}{x^2}\right)}^n$ is given by,

$T_{r+1}={}^{n}C_r{x}^{n-r}{\left(\frac{-3}{x^2}\right)}^r$

$={\left(-1\right)}^r\times {}^{n}C_r{3}^r x^{n-r-2r}$

$T_{r+1}={\left(-1\right)}^r\times {}^{n}C_r{3}^r x^{n-3r}...............\left(1\right)$

So, $T_1=T_{0+1}= {}^{n}C_0{3}^0 x^n=x^n$

$T_2=\left(-1\right)\times {}^{n}C_1{3}^1 x^{n-3}$

$T_3={}^{n}C_2{3}^2 x^{n-6}$

Now given sum is

$1- {}^{n}C_1·3+ {}^{n}C_2·3^2=376$

$\Rightarrow 1-3n+\frac{n\left(n-1\right)}{2}·9=376$

$\Rightarrow 1-3n+\frac{n^2-n}{2}·9=376$

$\Rightarrow 2-6n+9n^2-9n=752$

$\Rightarrow 9n^2-15n-750=0$

$\Rightarrow 3n^2-5n-250=0$

$\Rightarrow n=\frac{5\pm \sqrt{25+3000}}{6}$

$\Rightarrow n=\frac{5\pm 55}{6}$

$\Rightarrow n=10 \left\{\text{ignoring negative sign}\right\}$

Now, $T_{r+1}={\left(-1\right)}^r {}^{10}C_r 3^r x^{10-3r}$

So, for coefficient of $x^4$ we take 

$10-3r=4$

$\Rightarrow 3r=6$

$\Rightarrow r=2$

So, coefficient of $x^4$ is given by,

$T_{2+1}={\left(-1\right)}^2·{}^{10}C_2 3^2=45\times 9=405$.