1. ## Mathematical Induction

Question:

Use the principal of Mathematical Induction to prove 2|(n2+n) for all n>=0

2. Originally Posted by oldguy
Question:

Use the principal of Mathematical Induction to prove 2|(n2+n) for all n>=0
for n=0, $\displaystyle 2|(0^2 + 0) = 2|0$

suppose it is true that for n=k, $\displaystyle 2|(k^2 + k)$.
show that if n=k+1, then $\displaystyle 2|((k+1)^2 + (k+1))$.

now, $\displaystyle (k+1)^2 + (k+1) = k^2 + 2k + 1 + k + 1 = k^2 + k + 2k + 2 = k^2 + k + 2(k + 1)$..
notice that, $\displaystyle 2|k^2 + k$ and $\displaystyle 2|2(k + 1)$

therefore, $\displaystyle 2|k^2 + k + 2(k + 1)$ or $\displaystyle 2|(k+1)^2 + (k+1)$ . QED

3. Hello, oldguy!

Use Mathematical Induction to prove: .$\displaystyle 2\,|\,(n^2+n)$ for all $\displaystyle n \geq 0$
$\displaystyle S(n)\!: \;n^2+n$ is a multiple of 2.

Verify $\displaystyle S(1)\!:\;\;1^2 + 1 \:=\:2$ . . . True!

Assume $\displaystyle S(k)\!:\;\;k^2 + k \:=\:2a$ for some integer $\displaystyle a$

Add $\displaystyle 2k+2$ to both sides: .$\displaystyle k^2 + k + {\color{blue}2k + 2} \;=\;2a + {\color{blue}2k + 2}$

We have: .$\displaystyle k^2 + 2k + 1 + k + 1 \;=\;2a + 2k + 2$

. . . . . . . .. $\displaystyle \underbrace{(k+1)^2 + (k+1)}_{\text{Left side of S(k+1)}} \;=\;\underbrace{2(a + k + 1)}_{\text{multiple of 2}}$

Therefore, we have proved $\displaystyle S(k+1).$
. . The inductive proof is complete.

4. Thanks for the help.