We have recently been studying quadratic and cubic sequences in lessons by finding the 2nd and 3rd differences.
I noticed that for any sequence $an^x+bn^{x-1}$.... that $a=\frac{1}{x!}$ of the $x$th difference of the sequence $n^x$
Is there any explanation of this or a proof that this is always the case. I am only at GCSE level so it would be great if you could try to explain it using as little complex terminology and notation as possible.
Thank you!
The $x$-th difference of $n^x$ is $x!$ and the $x$-th difference of the other terms, $bn^{x-1}+\cdots+cn^0$ are zero.
If the leading term is $an^x$ with $a=\frac{1}{x!}$ then the $x$-th difference is $1$
And, yes there is a explanation and proof for this. The most easy, that i know, use something called "finite induction" (i don't know if you are familiar with this method of proof).
Informally, every time we apply the difference in a term of the form $n^x$ the result is something of the form $x(n^{x-1}+ \cdots$ other terms) (we decrease the degree of the polynomial by 1).
If we apply the difference one more time the result is $x(x-1) (n^{x-1}+\cdots$ other terms.
If we continue with this process in the $x-1$ difference we will come to $$ x(x-1)\cdots (2)(n+c)$$ , and one more time, using that the difference of $n$ is $1$ and constante is $0$, we obtain in the $x$-th aplication $$x(x-1)\cdots (2)(1)=x! $$ independent of $n$.