Question

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

Answer 1

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