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

Question

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$ = $k^2+2k+1$

Answer 1

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 $n=k+1$.

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[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.

Answer 2

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.

Answer 3

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.

Answer 4

Hint: $$(n+1)((n+1)+1)=(n+1)(n+2)=n(n+1)+2(n+1)$$

Answer 5

$n(n+1)$ 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.