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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?

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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 $$\begin{matrix} 1&&32&&243&&1024&&3125&&7776&&16807\\ &31&&211&&781&&2101&&4651&&9031\\ &&180&&570&&1320&&2550&&4380\\ &&&390&&750&&1230&&1830\\ &&&&360&&480&&600\\ &&&&&120&&120 \end{matrix}$$

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  • $\begingroup$ Your perspective is great... i simply atta k the problem directly without relying on its relationship to previously established facts..Any idea how will i start ur enrichment sir? $\endgroup$ – rosa Jan 21 at 14:47
  • $\begingroup$ I haven't proved it myself for now. I just observed it. $\endgroup$ – Larry Jan 21 at 15:00
  • $\begingroup$ Ah okay... ur great!!. Tnx! $\endgroup$ – rosa Jan 21 at 15:11
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    $\begingroup$ This is highly related to the fact that the $n$-th derivative of $x^n$ is $n!$. $\endgroup$ – Paul Sinclair Jan 21 at 15:51
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    $\begingroup$ How continuous subtraction connected to nth derivative? $\endgroup$ – rosa Jan 21 at 21:15
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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.

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