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.
You are watching: N(n+1)(n+2)/6
What am I doing wrong here?
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 $1=frac{1cdot2cdot3}{6}=1$, 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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