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

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- Mar 19th 2008, 04:36 PMdaaavoinduction
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 - Mar 19th 2008, 05:12 PMtopsquark
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