If 1+2+3+...+200=x, what is

Question

(1+2+3+...+200)-(2+4+6+...+200)?

(solution pls... tnx)

Answer 1

well,
(2+4+6+...+200) = 2(1+2+3+...+100)

1+2+3+...n = n(n+1)/2

So,

x = 200*201/2 and you have

200*201/2 - 2*100*101/2 = 200/2 (201-101) = 100*100 = 10,000

Or, you could note that

(1+2+3+...+200)-(2+4+6+...+200) = (1+3+5+...+199)

The sum of the first n odd numbers is n^2, so you have the first 100 odd numbers, summing to 100^2 = 10,000