summation - 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.

Answer

What you’ve really done up to this point is use your induction hypothesis to say that

$$(1)+(1+2)+\ldots+(1+2+\ldots+n)+\big(1+2+\ldots+n+(n+1)\big)$$

is equal to

$$\frac16n(n+1)(n+2)+\big(1+2+\ldots+n+(n+1)\big)\;.$$

To finish the induction step you must show that this quantity is equal to

$$\frac16(n+1)(n+2)(n+3)\;,$$

i.e., that

$$\frac16n(n+1)(n+2)+\big(1+2+\ldots+n+(n+1)\big)=\frac16(n+1)(n+2)(n+3)\;.\tag{1}$$

In order to do this, you need a nice closed form for the term

$$1+2+\ldots+n+(n+1)\;.$$

I’m sure that by this point you’ve learned a closed form for the sum of the first $m$ consecutive integers; substitute that (with $m=n+1$) for $1+2+\ldots+n+(n+1)$ on the lefthand side of $(1)$, and do some algebra to show that the quantity on the lefthand side then really does simplify to the quantity on the righthand side.