Find the sum of the series

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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.

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