Second difference on the sum of extreme values of square numbers I notice that in a series of consecutive square number $$1,4, 9, 16, 25, 36, 49, 64, 81, 100$$ if i add up the first element to the last element as well as the second element to the second to the last element i come up with the following result:$$101, 85, 73, 65, 61$$ the eventually get the absolute difference between 2 consecutive sum i have $$16, 12, 8, 4$$ by getting the second difference of this i got $$4,4,4,4$$ is this true to all $n^{2}?$
This pattern holds true even i do not start with $1^{2}$, that is for even number of squares, in case of odd number of squares i.e from $1^{2}$ to $9^{2}$, i can simply multiply the median, that is $5^{2}$, by two then do the same process the pattern still holds. I tried to prove this by letting $$n^{2}, (n +1)^{2}, (n + 2)^{2}, ..., (n+k)^{2}, (m - k)^{2}, . . .,(m - 1)^{2}, m^{2}$$ as i add both ends and perform subtraction among consecutive sums i got $$ (n+k)^{2} +(m - k)^{2} - 2((n+k)^{2} + (m - 1)^{2} + . . . +(n +1)^{2} +  (m - 1)^{2}) - n^{2} - m^{2}$$ but unfortunately i got stuck since I cant express sum of consecutive squares as single term,,, the internet say that its n(2n + 1)(n + 1)/6 but i cant connect this formula using expressions...any idea how to do this?
 A: 
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 
$$\begin{matrix}
1&&32&&243&&1024&&3125&&7776&&16807\\
&31&&211&&781&&2101&&4651&&9031\\
&&180&&570&&1320&&2550&&4380\\
&&&390&&750&&1230&&1830\\
&&&&360&&480&&600\\
&&&&&120&&120
\end{matrix}$$
A: 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.
