Why do I insert 2015 anywhere I like? #1

Let $S=\displaystyle \sum_{k=1}^{(n-1)/2} \sec\left(\dfrac{2\pi k}{n}\right)$ where $n$ is odd, $w=e^{2 \pi i / n}$ , $t=w+1/w=2\cos\left(\dfrac{2 \pi}{n}\right)$ and $P_k=w^k+1/w^k=2\cos\left(\dfrac{2 \pi k}{n}\right)$ .

Factoring $w^n=1$ we obtain: $w^{n-1}+w^{n-2}+\cdots+w+1=0$ , then divide by $w^{(n-1)/2}$ : $w^{(n-1)/2}+1/w^{(n-1)/2}+w^{(n-3)/2}+1/w^{(n-3)/2}+\cdots+w+1/w+1=0$ .

That expression is also $P_{(n-1)/2}+P_{(n-3)/2}+\cdots+P_1+1=0$ , which obviously factors as $\left(t-2\cos \dfrac{2 \pi}{n}\right)\left(t-2\cos \dfrac{4 \pi}{n}\right)\cdots\left(t-2\cos \dfrac{\pi(n-1)}{n}\right)$

Now, by Newton's sums we see that $P_1=t$ , $P_2=t^2-2$ and $P_k=t P_{k-1}-P_{k-2}$ . We only want to know the coefficient of $t$ and the constant term, so we see that the last term of $P_k$ is:

• $2(-1)^{k/2}$ if $k$ is even
• $k(-1)^{(k-1)/2}t$ if $k$ is odd

So, adding them:

$\cdots+\underbrace{(1-3+5-7+\cdots)}_{\left\lfloor \dfrac{n+1}{4} \right\rfloor \text{ terms}}t+\underbrace{(-2+2-2+\cdots)}_{\left\lfloor \dfrac{n-1}{4} \right\rfloor\text{ terms}}+1=0$

By Vieta's formulas, $S$ is $-2\times\dfrac{\text{coefficient of t}}{\text{constant term}}$ :

$S=-2\times\dfrac{1-3+5-7+\cdots}{(-2+2-2+\cdots)+1}$

After adjusting it, we end up with $S=2\left\lfloor \dfrac{n+1}{4}\right\rfloor(-1)^{(n-1)/2}$

In this problem $n=2015$ , so $S=\boxed{-1008}$ .

Let $z = e^{i\theta} = x+iy$ . Then:

$\cos((2n+1)\theta) = Re(z^{2n+1}) = \displaystyle \sum_{k=0}^n \binom{2n+1}{2k} x^{2(n-k)+1} (iy)^{2k}$ $= \displaystyle \sum_{k=0}^n (-1)^k \binom{2n+1}{2k} x^{2(n-k)+1} (1-x^2)^k$

which we denote as $f(x)$ . It's easy to see that:

$f(x) = a_{2n+1}x^{2n+1} + \ldots + a_1x^1 \quad ...(1)$

where

$a_1 = (-1)^n \binom{2n+1}{2n} = (-1)^n (2n+1) \quad ...(2)$

On the other hand, since $\cos((2n+1)\theta) = \dfrac12 \left( z^{2n+1} + \dfrac{1}{z^{2n+1}} \right) = 1 + \dfrac12 \left( z^{2n+1} + \dfrac{1}{z^{2n+1}} -2 \right)$

$= 1 + \dfrac{1}{2z^{2n+1}}(z^{2n+1} - 1)^2$ and

$(z^{2n+1} - 1)^2 = \displaystyle \prod_{k=0}^{2n} \left( z-e^{\frac{2k\pi i}{2n+1}} \right)^2 = \prod_{k=0}^{2n} \left(z-e^{\frac{2k\pi i}{2n+1}} \right) \left(z-e^{\frac{-2k\pi i}{2n+1}} \right)$

$= \displaystyle \prod_{k=0}^{2n} \left( z^2 + 1 - 2z \cos \left( \dfrac{2k\pi}{2n+1} \right) \right)$

We also have:

$f(x) = 1 + \dfrac12 \prod_{k=0}^{2n} \left( z + \dfrac1z - 2 \cos \left( \dfrac{2k\pi}{2n+1} \right) \right)$

$= 1 + 2^{2n} \prod_{k=0}^{2n} \left( x - 2 \cos \left( \dfrac{2k\pi}{2n+1} \right) \right) \quad ...(3)$

Now we compare the expressions $(1)$ and $(3)$ for $f(x)$ . We see from $(1)$ that the constant term in $(3)$ is zero. Therefore:

$2^{2n} \prod_{k=0}^{2n} \cos \left( \dfrac{2k\pi}{2n+1} \right) = 1$

Comparing the coefficients of $x$ in $(1)$ and $(3)$ , and dividing by the product above (which equals 1), we get:

$a_1 = \sum_{k=0}^{2n} \sec \left(\dfrac{2k\pi}{2n+1}\right) = 1 + 2\sum_{k=1}^{n} \sec \left(\dfrac{2k\pi}{2n+1}\right)$

Setting the above expression for $a_1$ equal to the expression in $(2)$ , we get:

$\sum_{k=1}^{n} \sec \left(\dfrac{2k\pi}{2n+1}\right) = \dfrac{(-1)^n(2n+1)-1}{2}$

which equals $n$ if $n$ is even, and $-(n+1)$ if $n$ is odd.

Since our $n$ for the problem is $n=1007$ , and since it's odd, the answer is $\boxed{-1008}$ .