Mathematical induction - how-to

I know it's not that hard, but I'm not getting it.

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

Does anybody know?

Three-fold path of Mathmatical Induction:
1. Show that statment is true for n = 1
2. Assume the statement is true for any positive integer, k.
3. Using your assumption in #2, prove that the statement is true for k+1.

1. n^2 + n = 1^2 + 1
= 2
2/2 = 1
Therefore, for n = 1, the statement is true.

2. Assume that 2 divides evenly into k^2 + k.

3.
(k + 1)^2 + (k + 1) = k^2 + 2k + 1 + k + 1
= k^2 + k + 2k + 2
= k^2 + k + 2(k + 1)

Since k^2 + k is divisible by 2, and 2(k+1) is also divisible by 2, the statement is true when n = k + 1. Thus, n^2 + n is evenly divisible by 2 whenever n is a positive integer.

For this induction we'll need to first prove the base case (n=1) and then show that if the proposition is true for some value of n, say, k, then it must be true for k+1.

Base case. (n=1)

n² + n = 1² + 1 = 2 Which clearly has 2 as a divisor, so we have proven the base case.

Inductive step. Assume that for some value of n, lets call it k, 2 divides k² + k

(k+1)² + (k+1)
= (k² + 2k + 1) + (k+1)
= (k² + k) + (2k + 2)
= (k² + k) + 2(k+1)

Now we assumed that 2 divides k² + k so lets say that k² + k = 2m (m must be an integer)

= 2m + 2(k+1)
= 2(m + k + 1) Which clearly has a factor of 2.

So we've shown that if the proposition is true for k, it is true for k+1 and since we've proved it is true for n = 1, the proposition is therefore true for all positive integers.

EDIT: Keith - the question told you to use induction so whilst your method is correct, it's not what was asked for.

Induction Proof :
Try for n = 1 ?
1^2 +1 = 1 + 1 = 2 yes , it divides by 2 ( evenly)