My favourite equation
1 + 2 + 3 + 4 = 4 * 5 / 2 = 10
1 + 2 + 3 + 4 + ... + n = n * (n + 1) / 2
The simplicity of the ``proof’’ is what I love about it. Let’s call the sum we are looking for \(x\):
1 + 2 + 3 + 4 + ... + n = x
# Now write the left-hand side again backwards
n + n-1 + n-2 + n-3 + ... + 1 = x
# Now add the two equations term-by-term
(n+1) + (n-1+2) + (n-2+3) + (n-3+4) + ... + (1+n) = 2x
(n+1) + (n+1) + (n+1) + (n+1) + ... + (1+n) = 2x
# There are n terms in the left-hand side, all of which are n+1
n * (n+1) = 2x
x = n * (n+1) / 2
