Constant Shenanigans!

Geometry Level 4

$\large \sum_{k=1}^{1000} \sqrt{\sqrt[\phi]{(k^{1 + \sqrt{5}})}\left(\dfrac{\left[(1+\sin(k\tau))(1-\sin(k\tau)\right]^{\left \lfloor \pi \right \rfloor}}{\left[\frac{1}{2}(e^{ik\tau} + e^{-ik\tau})\right]^{2\left \lfloor e \right \rfloor}}\right)} = \ ?$

Details and Assumptions :

$\tau$ is the almighty circle constant .
$\pi$ is the lesser circle constant.
$\phi$ is the golden ratio.
$e$ is the base of the natural logarithm.
$i$ is the imaginary unit.
$\lfloor x \rfloor$ is the greatest integer function.

The answer is 500500.

1 solution

Ryan Tamburrino
Mar 14, 2015

If anything, this problem is just a test of one's simplifying abilities. First, we'll focus on the first radicand.

$\sqrt[\phi]{k^{1+\sqrt{5}}} = k^{\dfrac{1+\sqrt{5}}{\phi}}$ Now, $\phi=\dfrac{1+\sqrt{5}}{2} \Rightarrow k^{\dfrac{1+\sqrt{5}}{\phi}} = k^2$

The second radicand is just composed with two expressions for $(\cos(k\tau))^2$ , and it reduces to

$\dfrac{(\cos(k\tau))^6}{(\cos(k\tau))^4} = (\cos(k\tau))^2$

So now all we need is

$\sum_{k=1}^{1000} k\cos(k\tau)$

Which is only $1+2+3+4+...+1000$ . Hence the answer $\dfrac{(1000)(1001)}{2} = \boxed{500500}$

edit in the last statement that it won't be 100 101/2 but 1000 1001/2

Writabrata Bhattacharya - 6 years, 3 months ago

Thanks, edited.

Ryan Tamburrino - 6 years, 3 months ago

Constant Shenanigans!

The answer is 500500.

1 solution

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