how would I go about proving:
the sum of the first n odd numbers is equal to n^2 for every n which is greater than or equal to 1.
Thanks
First comes the n = 1 case.
$\displaystyle \sum_{i = 1}^{1}(2i - 1) = 1 = 1^2$ (check!)
Now let the theorem be true for some n = k. We need to show that it is also true for n = k + 1.
$\displaystyle \sum_{i = 1}^{k + 1}(2i - 1) = \sum_{i = 1}^{k}(2i - 1) + (2(k + 1) - 1)$
By our assumption:
$\displaystyle \sum_{i = 1}^{k}(2i - 1) = k^2$
So
$\displaystyle \sum_{i = 1}^{k + 1}(2i - 1) = k^2 + 2(k + 1) - 1$
For your theorem to be true this must be equal to $\displaystyle (k + 1)^2$. Can you finish this?
-Dan