What does i+j=n-1 mean in this summation? $\sum_{i,j \in \mathbb{N} \ i+j=n-1}x^i y^j$

Question

What does i+j=n-1 mean in this summation? $$\sum_{\ \ \ i,j \in \mathbb{N} \\ i+j=n-1}x^i y^j$$

The result is $(x^{n-1}+x^{n-2}y+ \cdots+xy^{n-2}+y^{n-1})$

I can not relate $i+j=n-1$ to the eponents

Answer 1

We take the sum over all expressions of the form $x^iy^j$ where $i,j\in\Bbb N$ abd $i+j=n-1$. The pairs $(i,j)$ that match these conditions are $(n-1,0)$, $(n-2,1)$, and so on up to $(0,n-1)$. So we sum $x^{n-1}y^0+x^{n-2}y^1+\ldots +x^0y^{n-1}$.

Answer 2

It means you sum over all pairs of natural numbers $(i,j)$ such that $i+j = n-1$. Thus $i$ goes from $0$ to $n-1$ and $j = n-1-i$.

Answer 3

It is the sum of all ordered pairs $(i,j)\in \mathbb{N}^2$ such that $i+j=n-1$ in general we can write:

$$\sum_{\substack{i+j=n\\ (i,j)\in \mathbb{N}^2}}a_ib_j=\sum_{k=1}^{n-1}a_kb_{n-k}$$

The latter notation can be useful in a variety of contexts for instance when multiplying power series, (assuming convergence) we can pair up the elements based on what the sum of the indices is:

$$\left(\sum_{n=1}^\infty a_nx^n\right)\left(\sum_{n=1}^\infty b_nx^n\right)\left(\sum_{n=1}^\infty c_nx^n\right)\left(\sum_{n=1}^\infty d_nx^n\right)=\sum_{n=4}^{\infty}\left(\sum_{\substack{i+j+u+v=n\\ (i,j,u,v)\in \mathbb{N}^4}}a_ib_jc_ud_v\right)x^n$$

As you can see when summing over tuples with more then two elements, the latter notation ends up actually being a lot cleaner then was the case for just a two-tuple. You can read more about these sums here if you're interested: https://en.wikipedia.org/wiki/Cauchy_product