Xrhrft

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?

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 $left(n+k-1right)^2$, then the sum of these $k$ numbers is

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

This is

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

That should help you finish off the problem.