sum of series $(2^2-1)(6^2-1)+(4^2-1)(8^2-1)+\cdots \cdots +\cdots (100^2-1)(104^2-1)$ The sum of series $(2^2-1)(6^2-1)+(4^2-1)(8^2-1)+\cdots \cdots +\cdots (100^2-1)(104^2-1)$
Attempt Assume $\displaystyle S = \sum^{50}_{r=1}((2r)^2-1)((2r+4)^2-1) = \sum^{50}_{r=1}(4r^2-1)(4r^2+16r+15)$
$\displaystyle S = \sum^{50}_{r=1}(16r^4+64r^3+56r^2-16r-15)$
could some help me how to solve it, thanks 

You can find the sum using the idea of telescoping sum.
Since
$$((2r)^2-1)((2r+4)^2-1)=(2r-1)(2r+1)(2r+3)(2r+5)$$
we get
$$\small\begin{align}&\sum_{r=1}^{50}(2r-1)(2r+1)(2r+3)(2r+5)\\\\&=\frac{1}{10}\sum_{r=1}^{50}\color{red}{10}(2r-1)(2r+1)(2r+3)(2r+5)\\\\&=\frac{1}{10}\sum_{r=1}^{50}\color{red}{((2r+7)-(2r-3))}(2r-1)(2r+1)(2r+3)(2r+5)\\\\&=\frac{1}{10}\sum_{r=1}^{50}((2r+7)(2r+5)(2r+3)(2r+1)(2r-1)-(2r+5)(2r+3)(2r+1)(2r-1)(2r-3))\\\\&=\frac{1}{10}(107\cdot 105\cdot 103\cdot 101\cdot 99-7\cdot 5\cdot 3\cdot 1\cdot (-1))\end{align}$$
