0
$\begingroup$

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

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

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 $n!$.

$\endgroup$
  • $\begingroup$ ! could you prove it by induction please... $\endgroup$ – ATM Jan 17 '15 at 13:25
  • $\begingroup$ ! you have done, what I have given in my question. Could you help me to prove or prove my statement using Induction. Awaiting your response. $\endgroup$ – ATM Jan 17 '15 at 14:14
  • $\begingroup$ @ATM: Try it yourself first! You can edit your question and say what you've tried now. $\endgroup$ – user21820 Jan 18 '15 at 3:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.