<s>[INST] Mathematical Induction of $\frac{n(n-1)(n-2)}{6}$ 
Show that the number of triples that can be chosen from $n$ items is precisely $$\frac{n(n−1)(n−2)}{6}.$$

Suppose n = k+1,
We want $\frac{(k+1)k(k-1)}{6}$ therfore,
$\frac{k(k-1)(k-2)}{6}$ + (k+1)
and then solve the rest.
What am I doing wrong here?
 [/INST] I will show how to prove $$1+3+6+\cdots+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{6}.$$
For $n=1$ we have $1=\frac{1\cdot2\cdot3}{6}=1$, so the base case holds.
Assume for any $n-1$ we have that the sum holds: $$1+3+6+\cdots+\frac{(n-1)n}{2}=\frac{(n-1)n(n+1)}{6}.$$
We need to show that by adding the next term $\frac{n(n+1)}{2}$ to the left and right hand sides the equality still holds.  For the LHS it's obvious.  For the RHS, in detail:
$$\frac{(n-1)n(n+1)}{6}+\frac{n(n+1)}{2}=\frac{(n-1)n(n+1)+3n(n+1)}{6}=\frac{(n-1+3)n(n+1)}{6}=\frac{n(n+1)(n+2)}{6}.$$
Hopefully you see how to apply this to the proof you need.  (Hint: subtract 2 from each $n$.)
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