an arithmetic series with first term of 0 and a last term (200th)of 398.

formula for the sum of such a series is

S = n/2 [2a + (n-1)d]

a = first term,

d = common difference and

n = number of terms.

The sum for your case is therefore

S = 200/2 [2(0) + (200 – 1)2]

= 100 [0 + (199)2]

= 100 (398)

= 39800

Label the numbers in the series

0, 2, 4, 6, 8, …, 394, 396, 398.

1, 2, 3, 4, 5, …, 198, 199, 200.

Successive sums

up to numbers labelled 1, 2, 3, 4, 5

are 0, 2, 6, 12, 20

These sums can be factored into

1*0, 2*1, 3*2, 4*3, 5*4.

So sum to number labelled 200, ie 398 is

200*199 = 39800.

If you look at the set of numbers, you can add numbers on opposite sides to get a constant sum:

0 + 398 = 398

2 + 396 = 398

4 + 394 = 398

etc.

Because there are 200 even numbers, there will be 100 pairs.

So, 398 x 100 = 39,800

= 39800.

Just add pairs 0 and 398, 2 and 396, etc. They all add to 398. There are 100 pairs. . . .