Sum of the first $n$ triangular numbers - induction 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.

$$\frac{1}{6}n(n+1)(n+2)+(1+2+3+\cdots+n+1)=$$
$$=\frac{1}{6}n(n+1)(n+2)+\frac{(n+1)(n+2)}{2}=$$
$$=\frac{1}{6}n(n+1)(n+2)+\frac{3(n+1)(n+2)}{6}=$$
$$=\frac{1}{6}(n+1)(n+2)(n+3)$$
