Can anyone help me get started with this proof using induction? Thank a lot everyone.
Prove that the product of any three consecutive positive integers is divisible by 6.
Base case $\displaystyle 1 \times 2 \times 3 = 6$ so the proposition is true for the consecutive numbers $\displaystyle 1,\ 2,\ 3$.
Suppose the proposition is true for three consecutive numbers starting from $\displaystyle k$.
Then $\displaystyle 6 $ divides $\displaystyle [k \times (k+1) \times (k+2)] $, now consider the product of three consecutive integers starting for $\displaystyle k+1$:
$\displaystyle (k+1) \times (k+2) \times (k+3) = k \times (k+1) \times (k+2) + 3 \times (k+1) \times (k+2)$
but $\displaystyle 6 $ divides $\displaystyle [k \times (k+1) \times (k+2)] $ and as one of $\displaystyle (k+1)$ and $\displaystyle (k+2)$ must be even $\displaystyle 6 $ divides $\displaystyle [3 \times (k+1) \times (k+2)] $, and so we conclude that $\displaystyle 6 $ divides $\displaystyle [(k+1) \times (k+2) \times (k+3)] $.
Now we have proven the base case and that if the proposition is true for three consecutive integers starting from $\displaystyle k$ it is true for three consecutive integers starting from $\displaystyle k+1$, and so by induction it is true for every set of
three consective positive integers.
RonL
Hello, jzellt!
A slight variation of CaptainBlack's proof.
We want to proveProve that the product of any three consecutive positive integers is divisible by 6.
. . $\displaystyle \text{For all }n\in I^+,\;n(n+1)(n+2) \:=\:6a\;\;\text{for some integer }a.$
Verify $\displaystyle S(1)\!:\;\;1\cdot2\cdot3\:=\:6 $. . . True!
Assume $\displaystyle S(k)\!:\;\;k(k+1)(k+2) \;=\;6a$ for some integer $\displaystyle a$
Add $\displaystyle 3(k+1)(k+2)$ to both sides:
. . $\displaystyle k(k+1)(k+2) \:{\color{blue}+\:3(k+1)(k+2)} \;=\;6a \:{\color{blue}+ \:3(k+1)(k+2)}$
The left side is: .$\displaystyle k(k+1)(k+2)+ 3(k+1)(k+2)$
. . Factor: .$\displaystyle (k+1)(k+2)(k+3)\;\;{\color{red}\text{[L]}}$
The right side is: .$\displaystyle 3[2a + (k+1)(k+2)]$
. . $\displaystyle (k+1)(k+2)$ is the product of two consecutive integers.
. . Hence, it is divisible by 2:.$\displaystyle (k+1)(k+2) \:=\:2b $
The right side is:. $\displaystyle 3(2a+2b) \:=\:6(a+b)\;\;{\color{red}\text{[R]}}$
From [L] and [R],we have: .$\displaystyle (k+1)(k+2)(k+3)\:=\:6(a+b)$
We have established $\displaystyle S(k+1)$..The inductive proof is complete.