There are two ways to do it.  Here are both ways:
First way (without memorizing a frmula):

The 1st odd number is 1, 1 less than 2 which is twice 1.
The 2nd odd number is 3, 1 less than 4 which is twice 2.
The 3rd odd number is 5, 1 less than 6 which is twice 3.

So using that pattern

The 900th odd number is 1 less than twice 900, which is 1800, so the
900th odd number is 1799

Suppose the sum = N, then

 N =    1 +    3 +    5 +    7 + ··· + 1793 + 1795 + 1797 + 1799

N also equals that same sum with the numbers added in the reverse order:

 N = 1799 + 1797 + 1795 + 1793 + ··· +    7 +    5 +    3 +    1

Now let's write those two equations one under the other, and add equals 
to equals term by term:

 N =    1 +    3 +    5 +    7 + ··· + 1793 + 1795 + 1797 + 1799
 N = 1799 + 1797 + 1795 + 1793 + ··· +    7 +    5 +    3 +    1
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2N = 1800 + 1800 + 1800 + 1800 + ··· + 1800 + 1800 + 1800 + 1800

Since we know there are 900 terms on the right, that sum on the right is 
900 times 1800 or 1620000, so that equation bercomes

2N = 1620000

Dividing both sides by 2

 N = 810000

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The other way is to memorize the arithmetic series sum formula:

Sn = [2·a1 + (n-1)·d]

Where n=900, a1=1 because it is the 1st odd integer, and d=2,
because consecutive odd numbers differ by 2.
 
S900 = [2·(1) + (900-1)·(2)]

S900 = 450[2 + (899)·(2)]

S900 = 450[2 + 1798]

S900 = 450[1800]

S900 = 810000

Edwin