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How do you prove that

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

is divisible by 6 by using the method of mathematical induction?

According to my book $\begin{aligned} (n+1)(n+2)(n+3) &= n(n+1)(n+2)+3(n+1)(n+2)\\ &= 6k + 3*2k'\\ &= 6(k+k')\\ &=6k'' \end{aligned}$

But I wonder, where does that k come from anyway?

3 Answers 3

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The $k$ comes from the induction hypothesis: you are supposed to prove that $(n+1)(n+2)(n+3)$ is divisible by 6 assuming that $n(n+1)(n+2)$ is divisible by 6.

That's from where you get an integer $k$ for which $n(n+1)(n+2) = 6k.$

The $k'$ comes from the observation that $(n+1)(n+2)$ is the product of two subsequent numbers. One of which has to be even, so the product is even.

  • 1
    @MarcvanLeeuwen: mostly a tribute to how textbooks aren't very good at coming up with good examples of induction, I suspect :)2012-07-05
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Since one number out of every $2$ consecutive integers is divisible by $2$ and one out of every $3$ consecutive integers is divisible by $3$, therefore, in the product of $3$ consecutive integers $n(n+1)(n+2)$, atleast one number is divisible by $2$ and one divisible by $3$ and $\gcd(2,3)=1 $ $ \implies n(n+1)(n+2)$ is divisible by $2\times3=6$. $$ You can also do it this way: Since every integer is one of the forms: $6k,6k+1,6k+2,6k+3,6k+4 $ or $6k+5$, therefore, possibilities the product of three consecutive numbers is : 6k(6k+1)(6k+2)=6(k(6k+1)(6k+2))$ $(6k+1)(6k+2)(6k+3)=(6k+1)2(3k+1)3(2k+1)=6((6k+1)(3k+1)(2k+1))$ $(6k+2)(6k+3)(6k+4)=2(3k+1)3(2k+1)(6k+4)=6((3k+1)(2k+1)(6k+4))$ $(6k+3)(6k+4)(6k+5)=3(2k+1)2(3k+2)(6k+5)=6((2k+1)(3k+2)(6k+5)).$ Thus product of every possible combination of 3$ consecutive integers is coming out to be divisible by $6$.

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    :):):) me too..2012-07-05
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f(n)=n(n+1)(n+2)

Let f(n) is divisible by 6 be true for n=m.

f(m+1) = (m+1)(m+2)(m+3)=m(m+1)(m+2) + 3(m+1)(m+2) =f(m) + 3(m+1)(m+2). Now, product of two consecutive number is always even. So, the proposition is true for n = m+1 , if it is true for n=m.

By k means there exists an integer k such that n(n+1)(n+2)/6.