I need to prove that for all natural numbers $n$, $\frac{n(n+1)(n+2)}{6}$ is a natural number. The problem is that I can't seem to figure out what the series is, the LHS. Any help would be appreciated.

  • $\begingroup$ If you know binomial coefficients then you may notice that : $$\frac{n(n+1)(n+2)}{6}=\binom{n+2}{3}$$ which is a natural number . $\endgroup$ – user252450 Sep 9 '15 at 14:56

Hint: The numerator is a product of $3$ consecutive numbers, so we can say something about the prime factors of at least two of the numbers (think of the prime factors for $6$).

  • $\begingroup$ Sorry, but I don't understand your hint. $\endgroup$ – ematth7 Sep 9 '15 at 14:50
  • $\begingroup$ @ematth7 To start, if you know that you are multiplying n * (n+1), the result MUST be even because either n or n+1 has to be even. $\endgroup$ – Devon Parsons Sep 9 '15 at 14:52
  • $\begingroup$ @ematth7: As Devon has stated, exactly one of $n\cdot(n+1)$ will be even; in other words, there is a factor of $2$ in $n\cdot(n+1)$. Can we prove there must be a factor of $3$ in $n\cdot(n+1)\cdot(n+2)$? If so, then dividing by $6$ must yield a natural number. $\endgroup$ – Clayton Sep 9 '15 at 15:04

Base Case ($n = 1$): $\frac{1(1+1)(1+2)}{6}=1$ is true.

Inductive Hypothesis: Assume that for some k, it is true for n = k

Induction Step: $n = k + 1$

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

By Inductive Hypothesis, $\frac{k(k+1)(k+2)}{6}$ is a natural number.

If k is odd, $k+1$ is even
If k is even, $k+2$ is even

Since either $k+1$ or $k+2$ is even, $\frac{(k+1)(k+2)}{2}$ is a natural number

Since the sum of 2 natural numbers is a natural number,

$\frac{k(k+1)(k+2)}{6} + \frac{(k+1)(k+2)}{2}$ is a natural number

and we have proved the case for $n = k+1$


You seem to be completely misunderstanding the question, thinking, apparently, that the fraction you are given is the sum of some series. Whether it is or not is irrelevant to the question. You have n(n+1)(n+ 2).

Now, either n is even or n is odd. If n is even, it is divisible by 2- it can be written a n= 2k and so is n(n+ 1)(n+ 2)= 2k(2k+1)(2k+ 2) which is divisible by 2. If n is odd, it can be written n= 2k+ 1. n(n+1)(n+2)= (2k+1)(2k+1+1)(2k+1+2)= (2k+1)(2k+2)(2k+3)= 2(2k+1)(k+1)(2k+3). So what ever n is, n(n+ 1)(n+2) is divisible by 2.

Now, "3" is just a little harder. Either n is divisible by 3 or it has remainder 1 when divided by 3 or it has remainder 2 when divided by 3. If n is divisible by 3, n= 3k so n(n+1)(n+2)= 3k(3k+1)(3k+2) and is divisible by 3. If n has remainder 1 when divided by 3, it is of the form n= 3k+1 so n(n+1)(n+2)= (3k+1)(3k+1+1)(3k+1+2)= (3k+1)(3k+2)(3k+3)= 3(3k+1)(3k+2)(k+1) and is divisible by 3. If n has remainder 2 when divided by 3, it is of the form n= 3k+2 so n(n+1)(n+2)= (3k+2)(3k+2+1)(3k+2+2)= (3k+2)(3k+3)(3k+4)= 3(3k+2)(k+1)(3k+4) and is divisible by 3.

So for any integer n, n(n+1)(n+3) is divisible by both 2 and 3 and so divisible by 6.

  • $\begingroup$ Any reason why you didn't bother typesetting your question using MathJax or $\rm\LaTeX$? $\endgroup$ – Daniel W. Farlow Sep 9 '15 at 18:29

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