Proof by mathematical induction that $1+(1+2)+(1+2+3)+...+(1+2+3+...+n)=\frac{n(n+1)(n+2)}{6}$

Question

Prove by induction that for all positive integer values of $n$:

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

Please help with this proof by mathematical induction.

Answer 1

If you prove the hockey-stick identity by induction, the desired result is a corollary:$$\sum_{j=1}^n\sum_{k=1}^jj=\sum_{j=1}^n\binom{j+1}{2}=\binom{n+2}{3}=\frac{n(n+1)(n+2)}{6}.$$

Answer 2

If the $n$-th term is $a_n$, then $$ a_{n+1}=a_n+(1+2+\dots+n+(n+1))=a_n+\frac{(n+1)(n+2)}{2} $$ By the induction hypothesis, $$ a_{n+1}=\frac{n(n+1)(n+2)}{6}+\frac{(n+1)(n+2)}{2} $$ I guess you can finish.

Of course you need to know that $$ 1+2+\dots+n=\frac{n(n+1)}{2} $$ which you can prove by induction as well.

Answer 3

HINT.- From the given formula $$1 + (1 + 2) + (1 + 2 + 3) + \ldots + (1 + 2 + 3 + \ldots + n) = \frac{n(n + 1)(n + 2)}{6}$$ you have $\dfrac{(n-1)n(n+1)}{6}+(1+2+3+\cdots+n)$ so you can apply induction verifiant that $$\dfrac{(n-1)n(n+1)}{6}+\frac{n(n+1)}{2}$$ is equal to $\dfrac{n(n + 1)(n + 2)}{6}$ (obviously you have used the quite known formula for the sum of the first $n$ natural integers.