Coefficient of binomial expansion The coefficient of $x^3$ is $4$ times the coefficient of $x^2$ in the new expansion of $(1+x)^n$. Find the value of $n$.

Notice, we have
$$(1+x)^n=^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+\ldots +^nC_n.1^0.x^n$$  Given condition $$\color{red}{\text{coefficient of}\ x^3=4\times \text{coefficient of}\ x^2}$$ $$\implies ^nC_{3}=4\times ^nC_{2}$$ $$\frac{n!}{(n-3)!3!}=\frac{4\times n!}{(n-2)!2!}$$ $$\frac{1}{3}=\frac{4}{(n-2)}$$ $$n-2=12\implies n=14$$
