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I am aware that there is a formula to calculate the number of terms in a multinomial expression $(x_1+x_2+x_3+...x_r)^n$, i.e. $^{n+r-1}C_{r-1}$. However, this is in the case when the terms $x_1, x_2, x_3 ... x_r$ are different variables. In my case, the variables are the same; i.e. x, raised to different powers. Can someone please point me in the right direction?

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    $\begingroup$ I would hazard a guess that the terms that appear are the powers $x^k$ with $k$ ranging from $-2n$ to $2n$. At least when $n>1$. The coefficients will be more complicated. To get the idea why don't you work out small cases $n=2,n=3,\ldots$ by hand. $\endgroup$ Oct 9 '17 at 15:27
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    $\begingroup$ @JyrkiLahtonen That worked! Why don't you post an answer? That would be a proof by induction, right? $\endgroup$
    – Abhigyan
    Oct 10 '17 at 14:16
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HINT.-You have $$(x+\frac{1}{x}+x^2+\frac{1}{x^2})^n=\left(\frac{x^4+x^3+x+1}{x^2}\right)^n=\frac{(x+1)^n(x^3+1)^n}{x^{2n}}$$ Therefore you don't need in this case the multinomial

$$\displaystyle (x_1+x_2+x_3+x_4)^n=\sum_{1\le k_i\le 4}^{}\binom{n}{k_1,k_2,k_3,k_4}\prod_{i=1}^{4}x_i^{k_i}$$ which have $35$ terms. Anyway you have to calculate the corresponding simplifications in exponents

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We obtain \begin{align*} \color{blue}{\left(x+\frac{1}{x}+x^2+\frac{1}{x^2}\right)}&=\frac{1}{x^{2n}}(1+x+x^2+x^3)^n\\ &=\frac{1}{x^{2n}}\left(1+x+x^3\left(1+x\right)\right)^n\\ &=\frac{1}{x^{2n}}(1+x)^n(1+x^3)^n\\ &\color{blue}{=\frac{1}{x^{2n}}\sum_{j=0}^n\binom{n}{j}x^j\underbrace{\sum_{k=0}^n\binom{n}{k}x^{3k}}_{n+1\text{ terms}}}\tag{1} \end{align*}

Let's have a look at the three factors in (1).

  • The rightmost sum contains $n+1$ pairwise different terms with exponents $3k,0\leq k\leq n$.

  • The leftmost factor $\frac{1}{x^{2n}}$ does not change the number of different terms as it is just a shift of each exponent of $x$ by $-2n$.

  • Now we assume $n\geq 2$ and analyse the left sum. A multiplication with the first three terms $\binom{n}{0},\binom{n}{1}x,\binom{n}{2}x^2$ results in $3(n+1)$ terms \begin{align*} \sum_{j=0}^n a_kx^{3k\color{blue}{+0}},\sum_{j=0}^n b_kx^{3k\color{blue}{+1}},\sum_{j=0}^n c_kx^{3k\color{blue}{+2}} \end{align*} which gives a sum of increasing powers of $x^k, 0\leq k\leq 3(n+1)$. Additionally we observe that whenever $n(>2)$ is increased by $1$, the overall number of terms is increased by $1$.

We conclude: The number of different terms in the expansion of $\left(x+\frac{1}{x}+x^2+\frac{1}{x^2}\right)^n$ is \begin{align*} \color{blue}{1}&\qquad \color{blue}{n=0}\\ \color{blue}{4}&\qquad \color{blue}{n=1}\\ 3(n+1)+(n-2)=\color{blue}{4n+1}&\qquad \color{blue}{n\geq 2} \end{align*}

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