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 $n$.
 A: Solve $\binom{n}{3}=4*\binom{n}{2}$
A: The $r(0\le r\le n)$th term  $T_r$ in $(1+x)^n$ is $\binom nrx^r$
$$\implies\dfrac{T_{r+1}}{T_r}=\cdots=\dfrac{n-r}{r+1}$$
In our case,  $r=2$ and the ratio $=4$
A: coefficient of $x^3$ is four times the coefficient of $x^2$ 
$$(1+x)^n=\\\binom{n}{0}1^{n}x^{0}+\binom{n}{1}1^{n-1}x^{1}+{\color{DarkBlue} {\binom{n}{2}1^{n-2}x^{2} }}+{\color{Red}{\binom{n}{3}1^{n-3}x^{3}} }+...+\binom{n}{n}1^{n-n}x^{n}\\$$so 
$$\binom{n}{3}1^{n-3}=4*\binom{n}{2}1^{n-2}\\\binom{n}{3}=4\binom{n}{2}\\\frac{n(n-1)(n-2)}{3!}=4 \frac{n(n-1)}{2!} \\\frac{n-2}{6}=\frac{4}{2}\\n-2=12\\n=14$$ note that $$n(n-1) \neq 0 \rightarrow n \neq 0 ,n \neq 1$$ because if n=1 $(1+x)^1=1+x$ 
and if n=0  then $(1+x)^0=1$ these are not acceptable 
A: 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 $$\color{red}{\text{coefficient of}\ x^3=4\times \text{coefficient of}\ x^2}$$ $$\implies ^nC_{3}=4\times ^nC_{2}$$ $$\frac{n!}{(n-3)!3!}=\frac{4\times n!}{(n-2)!2!}$$ $$\frac{1}{3}=\frac{4}{(n-2)}$$ $$n-2=12\implies n=14$$
