Second difference on the sum of extreme values of square numbers

Lemma: First, realize that the difference of the difference between $3$ consecutive square numbers is $2$ $$\begin{matrix} 1&&4&&9&&16&&25&&36\\ &3&&5&&7&&9&&11\\ &&2&&2&&2&&2\\ \end{matrix}$$

Proof of lemma:

Suppose three consecutive square numbers $$n^2,~(n-1)^2,~(n-2)^2$$ Then $$n^2-(n-1)^2=2n-1$$ $$(n-1)^2-(n-2)^2=2n-3$$ $$[n^2-(n-1)^2]-[(n-1)^2-(n-2)^2]=(2n-1)-(2n-3)=2$$ Now to prove your result, suppose your series become $$n_1,n_2,n_3,\cdots,n_{k-1},n_{k}$$ What you are asking is to prove that $$[(n_{k}+n_1)-(n_{k-1}+n_2)]-[(n_{k-1}+n_2)-(n_{k-2}+n_3)]=4$$ Rewrite the equation and utilize the lemma, the result follows $$\underbrace{[(n_{k}-n_{k-1})-(n_{k-1}-n_{k-2})]}_{2}+\underbrace{[(n_{3}-n_2)-(n_{2}-n_1)]}_{2}=4$$ Surely this works for all $n^2$. I skipped some rigorous induction process.

Just for enrichment, the $n^{th}$ difference between $n+1$ consecutive numbers to the power of $n$ is $n!$.

For example,

$$\begin{matrix} 1&&8&&27&&64&&125&&216\\ &7&&19&&37&&61&&91\\ &&12&&18&&24&&30\\ &&&6&&6&&6 \end{matrix}$$ And $$\begin{matrix} 1&&32&&243&&1024&&3125&&7776&&16807\\ &31&&211&&781&&2101&&4651&&9031\\ &&180&&570&&1320&&2550&&4380\\ &&&390&&750&&1230&&1830\\ &&&&360&&480&&600\\ &&&&&120&&120 \end{matrix}$$

If you have $k$ consecutive squares starting from $n^2$ and going up to $\left(n+k-1\right)^2$, then the sum of these $k$ numbers is

$$\displaystyle\sum_{a=n}^{n+k-1} a^2 = \left(\displaystyle\sum_{a=1}^{n+k-1} a^2 \right) - \left(\displaystyle\sum_{a=1}^{n-1} a^2\right)$$

This is

$$\displaystyle\sum_{a=n}^{n+k-1} a^2\,\, = \,\,\frac{(n+k-1)(n+k)(2n+2k-1)}{6}-\frac{n(n-1)(2n-1)}{6}$$

That should help you finish off the problem.