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Prove with induction that $n(n+1)(n+2)$ is divisible by $6$ for all $n \ge 1$.

Test $n = 1$:

$$1(1+1)(1+2) = 6 $$

Hypothesis:

$$(\exists k \in \mathbb{N})(n(n+1)(n+2) = 6k)$$

This hypothesis is equivalent to

$$(\exists k \in \mathbb{N})(n^3+3n^2+2n = 6k)$$

Prove for $n+1$:

$$(n+1)(n+2)(n+3)$$

Expand it all and notice:

$$\color{blue}{n^3}+\color{blue}{3n^2}+3n^2+9n+\color{blue}{2n}+6$$

Replace with hypothesis:

$$\color{blue}{6k}+3n^2+9n+6$$

Maybe I can factorize that quadratic:

$$6k+(3n+6)(n+1)$$

If I could take $6$ out as a common factor for the whole expression, I would be done. But I am unsure on how to proceed here. How can I achieve that?

Saturn
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    A technical note, you should be showing that assuming the statement is true for $n$ implies the validity for $n+1$ not showing that $n+1$ can be worked out to be the assumed $n$ case! – mathematics2x2life Dec 03 '13 at 22:22
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    (Not by induction, but) $\displaystyle\binom{n+2}3=\frac{(n+2)(n+1)n}6$ is an integer, namely, the number of sets of size $3$ that we can form with elements coming from a pool of size $n+2$. – Andrés E. Caicedo Dec 03 '13 at 22:24
  • related: http://math.stackexchange.com/questions/533396/prove-by-mathematical-induction-that-n3-n-is-divisible-by-3-for-all-natur?rq=1 – abiessu Dec 03 '13 at 22:28

3 Answers3

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$(3n+6)(n+1)=3(n+2)(n+1)$

One of $n+1,n+2$ must be a multiple of 2(they are consecutive).

You have your factor of 6.

shyamupa
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Do you have to use induction?

The easiest proof is a direct proof, and involves casework on the remainders when $n$ is divided by $2$ and $3$.

Zubin Mukerjee
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  • Indeed, simple observations shows that one of the terms is divisible by $2$ and also that one of them is divisible by $3$, so it must be divisible by their product, which is $6$. – mathematics2x2life Dec 03 '13 at 22:23
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If you need to use induction, here's an approach that works: Induct on consecutive pairs of cases. I.e., check that $1\cdot2\cdot3$ and $2\cdot3\cdot4$ are both divisible by $6$ and then note that if you assume that $(n-1)n(n+1)$ and $n(n+1)(n+2)$ are both divisible by $6$, then you can conclude that $n(n+1)(n+2)$ and $(n+1)(n+2)(n+3)$ are both divisible by $6$ -- the first one obviously (it's part of the inductive hypothesis!), and the second because

$$(n+1)(n+2)(n+3) = (n-1)n(n+1)+6(n+1)^2$$

Barry Cipra
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