1+3+5+7 + ... + 1003 = what is the answer?

1 + (n-1) d = 1003

(n-1) 2 = 1002

n-1 = 501

n=502

Sum = 502 (1+1003) / 2 = 253,004

a) 1 + 2 + 3 + 4 + ... + 98

98 + 97 + 96 + 95 + ... + 1

sum 99 + 99 + 99 + 99 + ... + 99

there are 98 of these sums, the total is 98 * 99 = 9702, since this sum is twice the sum of the numbers 1 thru 98, we have

sum 1 + 2 + 3 + 4 + ... 98 = 9702/2 = 4851

b) 1 + 3 + 5 + 7 + ... + 1003

1003 + 1001 + 999 + 997 + ... + 1

sum 1004 + 1004 + 1004 + 1004 + ... 1004

There are 502 of these sums. the total is 502 * 1004 = 504008, since this sum is twice the sum of the numbers 1 + 3 + 5 + 7 + ... 1003, we have

sum 1 + 3 + 5 + 7 + ... + 1003 = 504008/2 = 252004