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I know you could expand it all out and take cases but is there a way to group it such that there is a general term? I've been trying to group it in order to establish a general term ie something like $((x + 1) + x^{-1})^4$ to make a binomial. Any hints?

user71207
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3 Answers3

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$$\left(x+\dfrac1x+1\right)^4=\sum_{r=0}^4\binom4r\left(x+\dfrac1x\right)^r$$

As the general term of $\left(x+\dfrac1x\right)^r$ is $$\binom rm x^{r-2m},$$

only the even $r$ will give $x$ independent terms

So, the required constant term $$=\binom40+\binom42\binom21+\binom44\binom42$$

lab bhattacharjee
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  • Wow, I was actually going to do this - a general term inside a general term! But I didn't because I assumed it would be wrong. Guess to have to be more confident – user71207 Jan 07 '21 at 07:19
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The contributions to the independent term are of the forms

  • $1,1,1,1$ ($1$ way)
  • $x,x^{-1},1,1$ ($4×3=12$ ways)
  • $x,x,x^{-1},x^{-1}$ ($\binom42=6$ ways)

Hence the independent term is $1+12+6=19$.

Parcly Taxel
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  • Thanks! But where did you get the ways from for the cases that involve $x$? – user71207 Jan 07 '21 at 07:21
  • @user71207 There must be as many $x$ terms as $x^{-1}$ terms. And the sum of $x$ and $x^{-1}$ terms cannot exceed four here. – Parcly Taxel Jan 07 '21 at 07:24
  • I mean where did you get $4\times 3$ and $4C2$ from? – user71207 Jan 07 '21 at 07:36
  • @user71207 For the second part: $4$ ways to choose the factor from which we choose $x$, $3$ ways after that for $x^{-1}$. For the third, we are choosing two factors from which we choose the $x$'s; the other two we choose $x^{-1}$'s. – Parcly Taxel Jan 07 '21 at 07:37
  • Sorry I really don't get it. What do you mean by "factor"? Perhaps you could show by writing it mathematically? – user71207 Jan 07 '21 at 08:29
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    @user71207 $$\binom4{0,0,4}+\binom4{1,1,2}+\binom4{2,2,0}=\frac{4!}{0!0!4!}+\frac{4!}{1!1!2!}+\frac{4!}{2!2!0!}==1+12+6=19$$ – bof Jan 07 '21 at 09:00
  • Thanks bof. I dont think @ParclyTaxel was actually using the multinomial theorem though but at least I see where they're coming from. Multinomial theorem seems like an easy alternative to general term - we can use multinomial theorem on binomials right? – user71207 Jan 07 '21 at 12:18
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An alternative way is to note that $\frac1x+1+x=\frac1x\cdot\frac{1-x^3}{1-x}$ so we are looking for the coefficient of $x^0$ in$$\frac1{x^4}\left(\frac{1-x^3}{1-x}\right)^4$$or the coefficient of $x^4$ in$$\begin{align*}\left(\frac{1-x^3}{1-x}\right)^4&=(1-x^3)^4(1-x)^{-4}\\&=\left[1+\binom41(-x^3)+...\right]\left[1+\binom{-4}1(-x)+...+\binom{-4}4(-x)^4+...\right]\end{align*}$$ which is $\binom{-4}4+\binom41\binom{-4}1=35-16=19$.

Shubham Johri
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