TITLE: Sum of the first $n$ triangular numbers - induction
QUESTION [6 upvotes]: Question:
Prove by mathematical induction that $$(1)+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)=\frac{1}{6}n(n+1)(n+2)$$ is true for all positive integers n.
Attempt:
I did the the induction steps and I got up to here:
$$RTP:\frac{1}{6}n(n+1)(n+2)+(1+2+3+\cdots+n+(n+1))=\frac{1}{6}(n+1)(n+2)(n+3)$$
Where do I go from here?
Thank you very much.

REPLY [2 votes]: The general equation for your expression is 
$$
S= \sum_{k=1}^{n}k(n-k+1)=n \sum_{k=1}^{n}k-\sum_{k=1}^{n}k^2 +\sum_{k=1}^{n}k
$$
what you need to know here are $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ and $\sum_{k=1}^{n}k^2 = \frac{n(n+1)(2n+1)}{6}$. 
You don't need induction to prove these, you may use the perturbation method from $\mathit{Concrete \ Mathematics}$, Chapter 2. I'll do the first sum, the second is similar.
$$
S_n +(n+1^2)=\sum_{k=1}^{n}k^2 + (n+1)^2 = 1 + \sum_{k=1}^{n}(k+1)^2 = 1+ S_n +2 \sum_{k=1}^{n}k+n
$$
Hence the $S_n$ cancel out, and you get the closed-form expression for $\sum_{k=1}^{n}k$
Once you get these, the rest is very easy:
$$
S= \frac{n^2(n+1)}{2}+\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}=\frac{n(n+1)(n+2)}{6}
$$