006 - sum of squares less square of sums

See Problem 6.

There are several stupid, straightforward ways of doing this. The following is one:

upper_bound = 100

sum_squares = 0
for i in xrange (1, upper_bound+1):
        sum_squares += i**2

square_sum = 0
for i in xrange (1, upper_bound+1):
        square_sum += i

square_sum *= square_sum

print sum_squares - square_sum

Execution time: < 1ms. xrange is used to decrease memory usage, and upper_bound is defined to make the solution generalisable. The solution could be physically shorter by combining the two loops, which shaves around a quarter of execution time.

Alternatively, we can calculate the sum of the sum of the terms without looping: if a series ranges from l to h, the average value of series member is (l+h)/2 and the sum of the series is thus n(l+h)/2. Further, the sum of an integer series from 1 to n can be calculated as n(n+1)(2n+1)/6. Thus:

upper_bound = 100

sum_squares = upper_bound * (upper_bound+1)*(2*upper_bound+1) / 6
square_sum = upper_bound * (1.0 + upper_bound) / 2.0
square_sum *= square_sum

print sum_squares - square_sum

which completes in half of the original. You could go even further and work out with a bit of algebra that the result should be -3n^4 - 2n^3 + 3n^2 + 2n.