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If you add numbers from 1 to n, the total is n(n+1)/2.

How do we know that? Because you can add n to 1, and (n-1) to 2 and so on, so you have n/2 values each totalling n+1 (it works even if n is odd, as you would have a half total left in the end).

Now, you want to add only the even integers, let's say from 1 to m.

You can say that m is the same as 2n throughout.

So the total of the first m even integers should be the same as twice the total of the first n integers, or n(n+1), and since we postulated that m=2n, the total is thus m/2(m/2+1).

Same way, if we want to add only the odd numbers, we can say that m is 2n-1 (if we said 2n+1, we'd miss 1 and we'd start at 3). Or, we can simply use the even total equation abive and subtract 1 for each value added up (so that 2 becomes 1, 4 becomes 3 and so on) that is to say n value less. If you work up the algebra, you end up with an interesting property: the equation reduces to a very neat ((m+1)/2)^2, that is the total of all the odd number from 1 to 7 is (8/2) ^2 or 16. Try it.

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13y ago

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