Can somebody help me with this mathematical induction proof?

Question

It's a recursive equation and I'm not sure how to demonstrate it through mathematical induction.

Answer 1

First you provisionally assume that the following statement is true for some $k$:

$a_k =2 + \frac{k(k+1)(2k+1)} {12} $

Then you use the recursion rule to prove (not assume!) that the next term follows the pattern, i.e.:

$a_{k+1} =2 + \frac{(k+1)(k+2)(2(k+1)+1)} {12} $

The algebra basically amounts to showing (this has to be shown!) that:

$2 + \frac{k(k+1)(2k+1)} {12} + \frac{(k+1)^2}2 = 2 + \frac{(k+1)(k+2)(2(k+1)+1)} {12} $

which is elementary algebra, so you should be able to handle this yourself.

And you only need to end with the base case, which is simply establishing that $a_0$ is indeed $2$ as stated, which is basically showing that $2 + \frac{(0)(0+1)(2(0)+1)} {12} = 2$, which is trivial.

If you cannot follow why this process works this way, you really should go back to your study materials and refresh yourself on mathematical induction. It helps to think of it as dominoes: you first show that if you pick a domino anywhere in a long (infinite) line and topple it, the next one will surely fall also. That establishes that all dominoes from that point on will just keep falling. Then you show the base case - that the first domino in the chain definitely falls to start with. Now you've proven that all dominoes definitely fall. That's the best elementary metaphor for mathematical induction I can come up with. Hope it helps.

Answer 2

Let $a_n$ be defined as in the problem, and for convenience let $f(n) = 2 + n(n+1)(2n+1)/12$. You're being asked to prove, by induction, that $a_n = f(n)$ for all $n\ge0$. By the definition of induction, that means proving the following two statements:

• $a_0=f(0)$ (this should be easy); and
• for every $n\ge1$, if $a_{n-1}=f(n-1)$, then $a_n=f(n)$.

Do you have an idea how to prove those statements?