prove that $n(n+1)$ is even using induction 
Prove that $n(n+1)$ is even using induction

The base case of $n=1$ gives us $2$ which is even.
Assuming $n=k$ is true, 
$n=(k+1)$ gives us $ k^2 +2k +k +2$ 
while $k(k+1) + (k+1)$ gives us $k^2+2k+1.$
whats is the next step to prove this by induction? I can't seem to show 
$ k^2 +2k +k +2$ = $k^2+2k+1$
 A: Hint:
$$(n+1)((n+1)+1)=(n+1)(n+2)=n(n+1)+2(n+1)$$
A: What you wrote in the second line is incorrect.
To show that $n(n+1)$ is even for all nonnegative integers $n$ by mathematical induction, you want to show that following:
Step 1. Show that for $n=0$, $n(n+1)$ is even;
Step 2. Assuming that for $n=k$, $n(n+1)$ is even, show that $n(n+1)$ is even for $n=k+1$. 

[Added:] In Step 2, what you really need to show is the following implication:

if $k(k+1)$ is even, then $(k+1)(k+2)$ is even. 

A: From $k^2$ + $3k$ + 2, you could do cases for k.
Case 1: If k is even, then let k = 2c for some integer c. Then you get $(2c)^2 $+ $3(2c)$ +2 , which could be written as 2(2cc) + 2(3c) +2 or 2(2cc + 3c + 1), which is even.
Case 2: If k is odd, then let k = 2c+1 for some integer c. Then you get $(2c+1)^2 $+ $3(2c+1)$ +2 , which could be written as $4c^2$ + $4c$ + 1 + $2c$+ + 1 + 2 , where you can simplify the expression down to 2 ($2c^2$ + $3c$ + 2). The factored out 2 indicates that it is even.
A: Claim: 
P(n): n(n+1) is even $\forall n \in \mathbb{N}$
Base case:
P(0): 0(1) = 0. Since 0 can be written in the form 2t, t $\in \mathbb{Z}$, 0 is even. The base case holds.
Alternative base case if you want to start at 1:
P(1): 1(2) = 2. Since 2 can be written in the form 2t, t $\in \mathbb{Z}$, 0 is even. The base case holds.
inductive hypothesis:
Suppose P(n) is true for some k $\in \mathbb{N}$. That is k(k+1) is even. Or $k^{2} + k$ is even.
inductive step:
P(k+1): 
$$(k+1)((k+1)+1) = (k+1)(k+2) $$
$$= k^{2} + 2k + k + 2$$
$$= (k^{2} + k) + (2k+2)$$
$$= (k^{2} + k) + 2(k+1) $$
By IH $k^{2} + k$ is even and since 2(k+1) is even by definition then P(k+1) is even. By principle of induction P(n) holds $\forall n \in \mathbb{N}$.
Note that it may require proof that even + even = even. Very simple proof however. 
A: $n(n+1)$ is an even number.
Take any $n\in\mathbb N$, then $n$ is either even or odd.

*

*Suppose $n$ is even $\implies n= 2m$ for some $m\implies n(n+ 1) = 2m(n+ 1)\implies n(n+ 1)$ is even.

*Suppose $n$ is odd $\implies n+1$ is even $\implies n+1 = 2m$ for some $m \implies n(n+1) = 2nm\implies n(n+1)$ is even.

