# Series ends with arithmetic series

Let us take series $a^2$. Where $a = 1, 2, 3, …$

The entire series of $a^2$ is look like: $A = 1, 4, 9, 16, …$

First step: The difference between every two terms of $A$ is: $B$ = $3, 5, 7, …$

Second step: The difference between $B$ is: $2, 2,…$ and looks like arithmetic progression.

In this problem, we have taken $a^2$ and within second step we have seen common difference and concluded this series as arithmetic series.

If you take $a^3$, within $3rd$ step we can conclude common difference is $3! = 6$ and arithmetic series.

My question is: How to prove the series $a^n$ for $a = 1, 2, 3,…$ gives common difference n! at $n$ th step, and gives arithmetic series. Also, I want to know any such similar observations are made so far.

Regards, ATM

• @vikrm! d=0 is not allowed – ATM Jan 17 '15 at 9:49
• @vikram! I mean, at second step you can see the common difference of B is 2. This is an A.P. series. – ATM Jan 17 '15 at 9:52

The operation you are doing at each step is just the forward difference operator (see Finite Difference at Wikipedia). In any case, you can prove your observation by induction on the step number. At step $0$ the function is a $n$-th degree polynomial with leading coefficient $1$. Then prove by induction that at step $t \in [0..n]$ the function is a $(n-t)$-th degree polynomial with leading coefficient $\prod_{k=1}^t (n+1-k)$. This would imply that at step $n$ the function is a $0$-th degree polynomial (which is a constant function) with leading coefficient $n!$.