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

The correct answer is 405
Given Binomial (x−x23​)n,x=0,n∈N,
Sum of coefficients of first three terms
nC0​−nC1​⋅3+nC2​32=376
⇒3n2−5n−250=0
⇒(n−10)(3n+25)=0
⇒n=10
Now general term 10Cr​x10−r(x2−3​)r
=10Cr​x10−r(−3)r⋅x−2r
=10Cr​(−3)r⋅x10−3r
Coefficient of x4⇒10−3r=4
⇒r=2
10C2​(−3)2=405