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?

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- Jul 20th 2010, 02:41 PMdwsmithInduction
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? - Jul 20th 2010, 02:50 PMeumyang
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}$ - Jul 20th 2010, 03:46 PMArchie Meade
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.