Series Summation: Solving for r^2 with (2r+1)^3 and (2r-1)^3 Equations

[spaghetti3451]

Homework Statement

Given that $$\left(2r + 1\right)^{3} - \left(2r - 1\right)^{3} = 24r^{2} + 2,$$
show that $$\sum r^{2} = \frac{1}{6}n(n+1)(2n+1).$$

Homework Equations

No idea!

The Attempt at a Solution

 

[eumyang]

Given this:
$$24r^{2} + 2 = \left(2r + 1\right)^{3} - \left(2r - 1\right)^{3}$$

For r = 1:
$$24(1)^{2} + 2 = \left(3\right)^{3} - \left(1\right)^{3}$$

Show us what the equation would look like if r = 2, 3, (n - 1), and n, exactly like I did for r = 1.

 

[Dick]

Homework Statement

Given that $$\left(2r + 1\right)^{3} - \left(2r - 1\right)^{3} = 24r^{2} + 2,$$
show that $$\sum r^{2} = \frac{1}{6}n(n+1)(2n+1).$$

Homework Equations

No idea!

The Attempt at a Solution

Sum both sides for r=1 to n. There's a lot of cancellation on the side with the cubes in it.

 

[Suk-Sci]

Its is just like showing the sum of first n²