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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 $n$.

Alex M.
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user258252
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    ,What is the relation between $x$ and $n$ – lab bhattacharjee Aug 01 '15 at 12:27
  • Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, make sure your question is well-posed: in the above, $x$ does not even appear in the quantity you want to analyze. – Clement C. Aug 01 '15 at 12:31

4 Answers4

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Solve $\binom{n}{3}=4*\binom{n}{2}$

John McGee
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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 $=4$

lab bhattacharjee
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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

Khosrotash
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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 $$\color{red}{\text{coefficient of}\ x^3=4\times \text{coefficient of}\ x^2}$$ $$\implies ^nC_{3}=4\times ^nC_{2}$$ $$\frac{n!}{(n-3)!3!}=\frac{4\times n!}{(n-2)!2!}$$ $$\frac{1}{3}=\frac{4}{(n-2)}$$ $$n-2=12\implies n=14$$

Harish Chandra Rajpoot
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