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I have the following question that I'm trying to work out an answer to:

"In the binomial expansion of $(1+x)^n$ where $n >= 4$, the coefficient of $x^4$ is $3/2$ times the sum of the coefficients of $x^2$ and $x^3$.

I have already figured it out in so far as getting $x$ by using $nCr$ for the coefficients like this:

$$\frac{3}{2}\left(\frac{n! }{(n-2)!2!} + \frac{n! }{(n-3)!3!}\right) = \frac{n! }{(n-4)!4!}$$

Apologies for the small brackets surroundig the left side of the equation - I am attempting to format correctly to little avail.

Regardless, I recognise that all that I need to do from here is solve for $n$. I am struggling to rearrange correctly, so would appreciate the help.

Kaciiey
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2 Answers2

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$$\frac{n(n-1)(n-2)(n-3)}{4!}=\frac32\left(\frac{n(n-1)(n-2)}{3!}+\frac{n(n-1)}{2!}\right)$$ or $$n^2-11n=0.$$

As you can check,

$$(x+1)^{11} \\=x^{11}+11x^{10}+55x^9+165x^8+330x^7+462x^6\\+462x^5+\color{green}{330}x^4+\color{green}{165}x^3+\color{green}{55}x^2+11x+1.$$

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We have$$\frac{n! }{(n-4)!4!}=\frac{3}{2}\left( \frac{n! }{(n-3)!3!}+\frac{n! }{(n-2)!2!}\right). $$ Simplify factorial ratios:$$\frac{n(n-1)(n-2)(n-3)}{4!}=\frac32\left(\frac{n(n-1)(n-2)}{3!}+\frac{n(n-1)}{2!}\right).$$ Divide by $n(n-1)$: $$\frac{(n-2)(n-3)}{4!}=\frac32\left(\frac{n-2}{3!}+\frac12\right)=\frac14(n-2)+\frac34.$$ Multiply by $24$:

$$n^2-5n+6=6(n-2)+18=6n-12+18=6n+6.$$

Add $-6n-6$:

$$n^2-11n=0.$$

Solutions of $n(n-11)=0$ are $n=0$ or $n=11$; given $n\ge4$, we take $n=11$.

J. W. Tanner
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