Proving that an expression divides a number 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?
 A: 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.
A: 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 $6$.
A: 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.
