1. ## Induction

Use induction to show that $\displaystyle 2|n(n+1)$

$\displaystyle n=1, \ 2|2$

Assume $\displaystyle p(k)$ is true.

$\displaystyle 2|(k+1)(k+2)\Rightarrow 2|(k^2+3k+2)$

$\displaystyle 2m=k(k+1)$

Now what?

2. Assume $\displaystyle n = k$ is true.
$\displaystyle \Rightarrow 2|(k)(k+1) \Rightarrow k(k + 1) = 2m$

Show true for n = k + 1:
\displaystyle \begin{aligned} (k + 1)(k + 2) &= k(k + 1) + 2(k + 1) \\ &= 2m + 2(k + 1) \\ &= 2(m + k + 1) \\ &\Rightarrow 2|(k + 1)(k + 2) \end{aligned}

3. Originally Posted by dwsmith
Use induction to show that $\displaystyle 2|n(n+1)$

$\displaystyle n=1, \ 2|2$

Assume $\displaystyle p(k)$ is true.

$\displaystyle 2|(k+1)(k+2)\Rightarrow 2|(k^2+3k+2)$

$\displaystyle 2m=k(k+1)$

Now what?
K or k+1 is even.
Any amount of an even value is even.

Via induction...

P(k)

k(k+1) is even ?

P(k+1)

(k+1)(k+2) is even ?

Try to prove that P(k) being true also causes P(k+1) to be true

Proof

$\displaystyle (k+2)(k+1)=k(k+1)+2(k+1)$

If P(k) is true, then P(k+1) is true as 2(k+1) is a multiple of 2 and so is even.