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?
[Math] The number of terms in the Multinomial Expansion $(x+\frac{1}{x}+x^2+\frac{1}{x^2})^n$
binomial theoremmultinomial-theorem
Best Answer
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 $1$.