TITLE: Induction and Recursion: $f(1)=2$ and $f(n)=f(n-1)+2n$ for $n>1$
QUESTION [5 upvotes]: I have bad time with induction and recursion and I have an exam soon.
We have this:
$$f(1)=2$$
$$∀n>1:f(n)=f(n-1)+2n$$
We need to proof that is the solution of f(n)=f(n-1)+2n,f(1)=2 is:
(the solution need be prooven with induction)
$$f(n)=n(n+1)$$
solution:
Base case:
n=1:
$$f(1)=2$$
$$f(1)=1(2)$$
We can assume that:
$$f(n)=f(n-1)+2n = f(n)n(n+1)$$
Inductive Step:
I don't know what to do next??
Any help will be appreciated!

REPLY [5 votes]: We have:
$$f(n)=f(n-1)+2n$$
And we need to prove:
$$f(n)=n(n+1)$$
Now let's check for $n=1$:
$$f(1)=1(1+1)=2$$
Now let's assume it's correct for n, i.e. that $f(n)=n(n+1)$, so we can use it.
$$f(n-1)+2n=n(n+1)$$
Now let's prove for $n+1$:
$$f(n+1-1)+2(n+1)=(n+1)(n+2)$$
$$f(n)+2n+2=(n+1)(n+2)$$
Now we can use the assumption, and substitute $f(n)$ with it's assumed value  $n(n+1)$:
$$n(n+1)+2n+2=(n+1)(n+2)$$
$$n^2+n+2n+2=n^2+3n+2$$
$$n^2+3n+2=n^2+3n+2$$
And that's correct.
Now by the Induction Axiom, we proved the theorem.