Generalization of the relation between square numbers and odd numbers

Question

We know that an easy way to obtain all the squares of the natural numbers is by adding starting from 1 all the natural odd numbers one at a time:

$1$ is itself a square

$1+3=4$

$1+3+5=9$

and so on...

I was wondering if this is just a specific case of a more generalized technique to obtain subsequentally and in order the terms of series of the form: $\sum_{n=1}^{\infty} {(x)^k}$, with $k$ being a non-zero natural number.

Answer 1

You probably mean something like this:

I liked to make tables like this in high school. Each new line below the line of $x^k$ is made of cells which are the difference between the top cell and the top-left cell.

The important thing is that the constant we obtain at the bottom is:

$$k!=2 \cdot 3 \cdot \ldots \cdot k$$

If you try to build a general relation from these tables, you obtain something like Newton's binomial:

$$(x+1)^k=\sum_{j=0}^k \left( \begin{array}( k \\ j \end{array} \right) x^j$$

Where:

$$\left( \begin{array}( k \\ j \end{array} \right)= \frac{k!}{j! (k-j)!}$$