Question

Let the sum of the coefficients of the first three terms in the expansion of ${\left(x-\frac{3}{x^2}\right)}^n,x\ne 0.n\in N$, be $376$. Then the coefficient of $x^4$ is ______

Answer 1

Answer: 405

Given Binomial ${\left(x-\frac{3}{x^2}\right)}^n,x\ne 0,n\in N$,
Sum of coefficients of first three terms
${}^n{C}_0-{}^n{C}_1\cdot 3+{}^n{C}_2{3}^2=376$
$\Rightarrow 3n^2-5n-250=0$
$\Rightarrow (n-10)(3n+25)=0$
$\Rightarrow n=10$
Now general term ${}^{10}{C}_r{x}^{10-r}{\left(\frac{-3}{x^2}\right)}^r$
$={}^{10}{C}_r{x}^{10-r}(-3{)}^r\cdot x^{-2r}$
$={}^{10}{C}_r(-3{)}^r\cdot x^{10-3r}$
Coefficient of $x^4\Rightarrow 10-3r=4$
$\Rightarrow r=2$
${}^{10}{C}_2(-3{)}^2=405$