Question

Prove that 2 divides $n^2+n$ whenever n is a positive integer.

Short answer

2 divides $n^2+n$ whenever n is a positive integer

Step-by-step solution

Step: 1

If n=1,

$1^2+1=2$

it is true for n=1.

Step: 2

Let P(k) be true.

$k^2+k$

We need to prove that P(k+1) is true.

Step: 3

$\begin{array}{r}(k+1{)}^2+(k+1)\\ =k^2+3k+2\\ =\left(k^2+k\right)+(2k+2)\\ =\left(k^2+k\right)+2(k+1)\end{array}$

It is true for P(k+1) is true.