What is the sum of 1, 3, 5, . . ., 97, 99 ?

The series 1, 3, 5, . . . can be described as 2n - 1 with n counting 1, 2, 3, 4, . . .

When we ask about the sum of 1, 3, 5, . . . up to 99, we're asking the same thing as the sum of 2n-1 from 1 up to 49 (note: 2n-1 when n is 49 makes 99).

You can see a demonstration (less wordy than my explanation below) by going here ( http://www.mathsisfun.com/numbers/sigma-calculator.html ) and entering 1 below the funny E, 49 above the funny E, and 2n-1 to the right of the funny E. The funny E is called a "summation" symbol.

The math way of writing the problem is:

Sum(i=1, n, 2i-1)

Note: i is the thing that counts 1, 2, 3, . . . and n is the final value that i counts up to. In this case n = 49.

Sum(i=1, n, 2i-1)

For the -1 part, we're just get -1 -1 -1 and so on, n times, for a total of -n. Let's leave that off for a moment.

Now we want the sum of 2i. The sum of 2i is just 2 * the sum of i.

We need a summation formula. From this page ( http://polysum.tripod.com/ ) we get:

Sum from i = 1 to n is n(n+1)/2

But we want the sum of 2i, so it's 2n(n+1)/2 or just n(n+1)

So. . . the sum of 2i - 1 from 1 to n is:

n(n+1) - n

But our n = 49, so:

49(49+1) - 49

49(50) - 49

2450 - 49

2401