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What is the sum of 1+3+5+7+9+....+1993+1995+1997+1999?
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Here's a way for you to think of this problem. 1+3+5+7+9+....+1993+1995+1997+1999 = = (1 + 1999) + (3 + 1997) + (5 + 1995) + (7 + 1993) + ...
If it were 1 + 2 + 3...+n, the sum would be (1/2)(n)(n+1) Perhaps you can do something with that. Yes, take that sum and subtract the even ones, 2 + 4 + 6 +...n and you know that that even sum is twice the sum from 1 to n/2, So, total sum 1 to n = even sum + odd sum odd sum = total - even That should work. 1 + 3 + 5 + = 1+2+3+4+5...- 2+4+...
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