difference of difference is constant question

Question

Why is the difference of the differences of squares is 2? e.g.: 4, 9, 16, 25, 36 ... are all squares, the difference of two adjacent are: 5, 7, 9, 11 ... and the difference of that is always 2, why is that happening?

thanks

Answer 1

Your function is $f(n)=n^2$, the difference is $$f(n+1)-f(n)=(n+1)^2-n^2=n^2+2n+1-n^2=2n+1:=g(n)$$

Now if you take the next difference it is $$g(n+1)-g(n)=2(n+1)+1-(2n+1)=2$$

This kind of thing you can find in calculus of finite differences.

We can define an operator $\Delta$, such that $\Delta f(n)=f(n+1)-f(n)$

and applying "m" times $$ \Delta^m f(n).$$ If we have $f(n)$ polynomial of degree "m", we can show in general $$\Delta^m f(n)=C $$ where $C$ is a constant.

Edit: We can show that $C= m! a_m$ where $f(x)=a_mx^m+\ldots+a_0x^0$

In your case $f(x)=x^2$ then $\Delta^2 x^2=C=2!.1=2.$

Answer 2

Series of squares is $(x-1)^2, x^2, (x+1)^2$.

Series of difference of squares is $x^2-(x-1)^2=2x-1, (x+1)^2-x^2=2x+1$.

and series of difference of difference is $2x+1-(2x-1)=2$.