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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

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    The sums $\sum_{r = 1}^{50}r, \sum_{r = 1}^{50}r^2, \sum_{r = 1}^{50}r^3$ and $ \sum_{r = 1}^{50}r^4$ may be solved separately. – 2017-01-16
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    @Arther how i solve $\sum^{50}_{r=1}r^4,$ please explain me – 2017-01-16
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    I googled and found [this](http://www.maa.org/press/periodicals/convergence/sums-of-powers-of-positive-integers-pierre-de-fermat-1601-1665-france), which looks like it might be helpful. – 2017-01-16

2 Answers 2

10

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 $$\small\begin{align}&\sum_{r=1}^{50}(2r-1)(2r+1)(2r+3)(2r+5)\\\\&=\frac{1}{10}\sum_{r=1}^{50}\color{red}{10}(2r-1)(2r+1)(2r+3)(2r+5)\\\\&=\frac{1}{10}\sum_{r=1}^{50}\color{red}{((2r+7)-(2r-3))}(2r-1)(2r+1)(2r+3)(2r+5)\\\\&=\frac{1}{10}\sum_{r=1}^{50}((2r+7)(2r+5)(2r+3)(2r+1)(2r-1)-(2r+5)(2r+3)(2r+1)(2r-1)(2r-3))\\\\&=\frac{1}{10}(107\cdot 105\cdot 103\cdot 101\cdot 99-7\cdot 5\cdot 3\cdot 1\cdot (-1))\end{align}$$

1

Being lazy, we could use the classical sums of powers of integer numbers as given here. This would lead to

$$ S_n = \sum^{n}_{r=1}(16r^4+64r^3+56r^2-16r-15)=\frac{1}{5} \left(16 n^5+120 n^4+280 n^3+180 n^2-71 n\right)$$

For sure, much more elegant is mathlove's solution.