The cosine summation

Geometry Level 4

The value of $\displaystyle\sum_{r=0}^{12} \binom{12}{r} \cos \frac {r\pi}{6}$

is of the form $-(x+\sqrt {y})^{z}$ , where $x,y$ and $z$ are integers, and $z$ is as large as possible.

Find $x+y+z$ .

1 solution

Pratik Shastri
May 29, 2014

$\displaystyle\sum_{r=0}^{12} \binom{12}{r} \times cos \frac{r\pi}{6}$

Can be rewritten as the real part of the expression $E=\displaystyle\sum_{r=0}^{12} \binom{12}{r} \times e^{\frac{ir\pi}{6}}$

$E=\displaystyle\sum_{r=0}^{12} \binom{12}{r} \times {\left[e^{\frac{i\pi}{6}}\right]}^{r} \times 1^{12-r}$

So, $E=\left[1+e^{\frac {i\pi}{6}}\right]^{12}$

$=\left[1+\frac{\sqrt {3}}{2}+\frac{i}{2}\right]^{12}$

Now, let $z= 1+\frac {\sqrt 3}{2}+\frac{i}{2}$

$|z|=\sqrt {2+\sqrt 3}$

And $arg (z)=\frac {\pi}{12}$

So, $E=\left[\sqrt {2+\sqrt 3} \times e^{\frac{i\pi}{12}}\right]^{12}........z=|z|e^{i \times arg (z)}$

$Re (E)= Re\left[({2+\sqrt 3})^{6} \times e^{i\pi}\right]........(e^{i\pi}=-1)$

$Re(E)=\boxed{-(2+\sqrt 3)^{6}}----------Ans$

So, $x+y+z=\boxed{11}$

As there is no constraint on the values of x, y and z; there may be more than one answer exists.

sujoy roy - 7 years ago

That's true. I've updated the question to make $z$ as large as possible, which will make the answer unique.

Calvin Lin Staff - 7 years ago

That's where I lost it!! It got the form of Re[{2+sqrt(3)+i}/(2)]^(12) and then w/a it... It was a long sequence and so I took the value of z =12

Kunal Gupta - 6 years, 8 months ago