A pattern on finding squares for which their sum is $\Big(\sum\limits_{i=0}^j x^i\Big)^2$

Question

Apologies for the inactivity. I haven't been doing so well in life, lately, but I'm glad to be back at some maths! Here's a pattern I discovered. I'm not good at explaining with words, so hope you get it. It shouldn't be too difficult. Can it be proven?

For all $x\in\mathbb{R}$,

$$\begin{align}\bigg(\sum_{i=0}^1 x^i\bigg)^2&=x^2+2x+1\\ &=(x+1)^2\end{align}$$

$$\begin{align}\bigg(\sum_{i=0}^3 x^i\bigg)^2&=x^6+2x^5+3x^4+4x^3+3x^2+2x+1\\ &=(x+1)^2+2(x^2+x)^2+(x^3+x^2)^2\end{align}$$

$$\begin{align}\bigg(\sum_{i=0}^5 x^i\bigg)^2&= x^{10} +2x^9+3x^8+4x^7+5x^6+6x^5+5x^4+4x^3+3x^2+2x+1\\ &=(x+1)^2+2(x^2+x)^2+3(x^3+x^2)^2+2(x^4+x^3)^2+(x^5+x^4)^2\end{align}$$

Clearly there is a pattern with the exponents in the square of each partial sum, as well as the coefficients. Can one prove this pattern continues ad infinitum?

For $\sum\limits^j$ I think the pattern also applies to even $j$ and not just odd $j$ as shown above, but I wrote this all on a napkin as I was eating at a restaurant, and now I've lost the napkin. (True story.)

Any help would be much appreciated. Thank you :)

Answer 1

Induction.

We want to show that $$\left(\sum\limits_{i=0}^{2k+1}x^i\right)^2=(k+1)(x^{k+1}+x^k)^2+\sum_{j=1}^kj[(x^j+x^{j-1})^2+(x^{2k+2-j}+x^{2k+1-j})^2]$$ for all $k\ge1$. As you have shown the base case, suppose this is true for some $k\in\Bbb N$. Then consider $k+1$. \begin{align}\left(\sum\limits_{i=0}^{2k+3}x^i\right)^2&=\left(x^{2k+3}+x^{2k+2}+\sum\limits_{i=0}^{2k+1}x^i\right)^2\\&=(x^{2k+3}+x^{2k+2})^2+2(x^{2k+3}+x^{2k+2})\sum\limits_{i=0}^{2k+1}x^i+\left(\sum\limits_{i=0}^{2k+1}x^i\right)^2\\&=(x^{2k+3}+x^{2k+2})^2+x^{2k+2}+x^{4k+4}+2\sum_{i=2k+2}^{4k+3}x^i\\&\quad+(k+1)(x^{k+1}+x^k)^2+\sum_{j=1}^kj[(x^j+x^{j-1})^2+(x^{2k+2-j}+x^{2k+1-j})^2]\end{align} and we want to show that $$\left(\sum\limits_{i=0}^{2k+3}x^i\right)^2=(k+2)(x^{k+2}+x^{k+1})^2+\sum_{j=1}^{k+1}j[(x^j+x^{j-1})^2+(x^{2k+2-j}+x^{2k+1-j})^2].$$ Subtracting the two equations yields $$\sum_{i=k+2}^{2k+2}(x^i+x^{i-1})^2=x^{2k+2}+x^{4k+4}+2\sum_{i=2k+2}^{4k+3}x^i$$ which is true. $\square$

Answer 2

I would find the closed form of the RHS (and compare it to the LHS). For $j=2n-1$, it is equal to $$\sum_{k=1}^{n}k(x^k+x^{k-1})^2+\sum_{k=1}^{n-1}k(x^{2n-k}+x^{2n-k-1})^2\\=(x+1)^2\left(f_n(x^2)+x^{4(n-1)}f_{n-1}(1/x^2)\right),$$ where $$f_n(y)=\sum_{k=1}^{n}ky^{k-1}=\frac{d}{dy}\sum_{k=0}^{n}y^k=\frac{ny^{n+1}-(n+1)y^n+1}{(y-1)^2}.$$ Substitution gives $\left((x^{2n}-1)/(x-1)\right)^2$ as expected. For even $j$, there's a difference - find it out.