Each number in the sum is even, so we can remove a factor of 2.
2 + 4 + 6 + 8 + ... + 78 + 80 = 2 (1 + 2 + 3 + 4 + ... + 39 + 40)
Use whatever technique you used in (a) and (b) to compute the sum
1 + 2 + 3 + 4 + ... + 39 + 40
With Gauss's method, for instance, we have
S = 1 + 2 + 3 + ... + 38 + 39 + 40
S = 40 + 39 + 38 + ... + 3 + 2 + 1
2S = (1 + 40) + (2 + 39) + ... + (39 + 2) + (40 + 1) = 40×41
S = 20×21 = 420
Then the sum you want is 2×420 = 840.
