2+4(3²)+6(5²)+...+2n(2n-1)² is the sum to n terms.
The first term (n=1) can be rewritten 2(1²).
If we expand the nth term we get:
2n(4n²-4n+1)=8n³-8n²+2n.
From this expansion we only need the sum to n terms the individual sums of the natural numbers, the squares and the cubes of the natural numbers.
The sum of the first n natural numbers is n(n+1)/2:
1, 3, 6, 10, 15, 21,...
The sum of the cubes:
1, 9, 36, 100, 225,...
But these are squares of:
1, 3, 6, 10, 15,...
Therefore the sum of cubes is (n(n+1)/2)².
Without offering proof, the sum of the squares of the natural numbers is:
n(n+1)(2n+1)/6.
So now we have enough to find the sum to n terms of the given series:
8(n(n+1)/2)²-4n(n+1)(2n+1)/3+n(n+1).
For the given series n=11, so the sum is:
8(66)²-44(12)(23)/3+132=30932.