Proof of $\frac{n(n-1)(n-2)}{6}=1+3+6+10...$

Question

Proof of title(implies proof of this type of sequence) Also how does this link to further proofs of $\frac{n(n-1)(n-2)(n-3)}{24}$ etc. and the proof of $\frac{n(n-1)}{2}$.

See also Sum of the first $n$ triangular numbers - induction and the posts linked there. (And probably there are some other posts about the same sum, this was one I was able to find relatively quickly.) — Martin Sleziak, Jun 02 '17 at 07:23

score 2 · Answer 1 · answered Jun 02 '17 at 05:43

2

These are just Hockey-sticky identities. Left side is precisely ${n}\choose{3}$, right side is $\sum {{n-1}\choose{2}}$ and the identity itself is easily deduced from recursive definition of binomial coefficients.

answered Jun 02 '17 at 05:43

Chen

78

score 1 · Answer 2 · answered Jun 02 '17 at 05:22

1

HINT:

$$\sum_{r=1}^n\dfrac{r(r+1)}2=\dfrac12\sum_{r=1}^nr^2+\dfrac12\sum_{r=1}^nr$$

Now use this.

answered Jun 02 '17 at 05:22

lab bhattacharjee

274,582

Observe that $$3-1=2,6-3=3,10-6=4$$ etc. So, the $r$th term of $1,3,6,10$ will be $$1+2+3+4+\cdots+r=?$$ – lab bhattacharjee Jun 02 '17 at 05:30

Proof of $\frac{n(n-1)(n-2)}{6}=1+3+6+10...$

2 Answers2