Quadratic Sum mini algebra

If $x^{2}-x+1=0$ . Then by quadratic formula $x= \frac{1}{2} + i\frac{\sqrt3}{2}$ or $x= \frac{1}{2} - i\frac{\sqrt3}{2}$ .

$(x-\frac{1}{x})^{2}+( x+\frac{1}{x} )^{2}+( x^{2}+\frac{1}{x^{2}} )^{2}+ ( x^{3}+\frac{1}{x^{3}} )^{2}+...+( x^{2015}+\frac{1}{x^{2015}} )^{2}$

$=$ $( x-\frac{1}{x})^{2}$ + $\sum_{n=1}^{2015}{ ( x^{n}+\frac{1}{x^{n}})^{2}}$ .

Let $x=\cos 60^\circ +i\sin60^\circ$ .

$( \cos 60^\circ +i\sin60^\circ-\frac{1}{\cos 60^\circ +i\sin60^\circ})^{2}$ + $\sum_{n=1}^{2015}{((\cos 60^\circ +i\sin60^\circ)^{n}+\frac{1}{(\cos60^\circ +i\sin60^\circ)^{n}})^{2}}$ $=$

$( \cos 60^\circ +i\sin60^\circ-(\cos 60^\circ -i\sin60^\circ))^{2}$ + $\sum_{n=1}^{2015}{(\cos (60n)^\circ +i\sin (60n)^\circ+\frac{1}{\cos (60n)^\circ +i\sin (60n)^\circ})^{2}}$ $=$

$(2i\sin60^\circ)^{2} + \sum_{n=1}^{2015}{(\cos (60n)^\circ +i\sin (60n)^\circ+ \cos (60n)^\circ -i\sin (60n)^\circ)^{2}} =$

$-4\sin^2 60^\circ$ + $\sum_{n=1}^{2015}{(2\cos (60n)^\circ )^{2} }=$

$-4(\frac{\sqrt3}{2})^{2} + \sum_{n=1}^{2015}{4\cos^2 (60n)^\circ}=$

$-3$ + $4\sum_{n=1}^{2015}{\cos^2 (60n)^\circ}=$ .

Cosine is a cyclic function so the values of ${\cos (60n)^\circ}$ repeat every six times because $\frac{360}{60}=6$ . 2015 mod 6 is 5 so the sum of the first six terms of the original summation repeats 335 times with 5 left over or 336 times minus the last value. The equation then becomes.

$-3$ $-\cos^2 (360)^\circ$ + $4\sum_{n=1}^{336}{\cos^2 (60)^\circ+\cos^2 (120)^\circ+\cos^2 (180)^\circ+\cos^2 (240)^\circ+\cos^2 (300)^\circ+\cos^2 (360)^\circ }$ =

$-3$ - $1^{2}$ + $4\sum_{n=1}^{336}{(\frac{1}{2})^{2}+(\frac{-1}{2})^{2}+ (-1)^{2}+(\frac{-1}{2})^{2}+(\frac{1}{2})^{2}+ (1)^{2} } =$

$-7$ + $4\sum_{n=1}^{336}{3}= -7+12\sum_{n=1}^{336}{1}=-7+12*336 =\boxed{4025}.$

note that for $x \ne -1$

we have $x^{3}-x^{2}+x = 0$ and $x^{2}-x+1 = 0$

sum that both

$x^{3} +1 = 0$ or $x^{3} = -1$ for $x \ne -1$

then note that $x \ne 0$ divide that LHS and RHS by $x$

we have $x+\frac{1}{x} = 1$

squared that LHS and RHS also we have $x^{2} +\frac{1}{x^{2}} = -1$

$$ $\,\,\, \,\,\,\,\, \,\,\,\,\, \bullet$ case $1$

$\left ( x - \frac{1}{x} \right )^{2} = x^{2} + \frac{1}{x^{2}} -2 = -1 -2 = -3$

$$ $\,\,\, \,\,\,\,\, \,\,\,\,\, \bullet$ case $2$

$\left ( x^{3k} +\frac{1}{x^{3k}} \right )^{2} = \left ( (x^{3})^{k} +\frac{1}{(x^{3})^{k}} \right )^{2} = \left ( 1+1 \right )^{2} = 4$ for some $k \in \mathbb{N}$ $$

$\left ( x^{3k+1} +\frac{1}{x^{3k+1}} \right )^{2} = \left ( x(x^{3})^{k} +\frac{1}{x(x^{3})^{k}} \right )^{2}$ $= \left ( x +\frac{1}{x} \right )^{2} = x^{2}+\frac{1}{x^{2}}+2 = -1+2 = 1$ for some $k \in \mathbb{N}\cup \left \{ 0 \right \}$ $$

$\left ( x^{3k+2} +\frac{1}{x^{3k+2}} \right )^{2} = \left ( x^{2}(x^{3})^{k} +\frac{1}{x^{2}(x^{3})^{k}} \right )^{2} = \left ( x^{2} +\frac{1}{x^{2}} \right )^{2} = (-1)^{2} = 1$ for some $k \in \mathbb{N}\cup \left \{ 0 \right \}$ $$ then $\left ( x-\frac{1}{x} \right )^{2}+\left ( x+\frac{1}{x} \right )^{2}+\left ( x^{2}+\frac{1}{x^{2}} \right )^{2}+\left ( x^{3}+\frac{1}{x^{3}} \right )^{2}+...+\left ( x^{2015}+\frac{1}{x^{2015}} \right )^{2}$

$= -3 + 4\left \lfloor \frac{2015}{3} \right \rfloor + 1\left ( 2015 - \left \lfloor \frac{2015}{3} \right \rfloor \right ) = \boxed{4025}$

since x is -w or -w^2 putting x as -w and expanding all brackets and forming GP we get the required sum as 4030-2-3=4025

note: w is cube root of unity

Consider $f(x)=x^3+1$ .

Then we have that $\frac{x^3+1}{x+1}=x^2-x+1$ and so the roots of $x^2-x+1$ are the complex roots of $x^3+1$ which are $\omega_{1}=e^{i\frac{\pi}{3}}$ and $\omega_{2}=e^{-i\frac{\pi}{3}}=\frac{1}{\omega_{1}}$ .

We have to evaluate the expression $(x-\frac{1}{x})^2+\sum_{a=1}^{2015} (x^{a}+\frac{1}{x^a})^2$ for $x=\omega_{1}$ or $x=\omega_{2}$ ,

but it is clear this expression does not change if we exchange $\omega_{1}$ and $\omega_{2}$ and so we could evaluate the expression just for $x=\omega_{1}$ .

The first term is equal to $(2*\Im{e^{i\frac{pi}{3}}})^2=(2\sin(\frac{pi}{3}))^2=(i\sqrt(3))^2=-3$ .

For the second term, note that we could rewrite it as

$\sum_{a=1}^{2015} \omega_{1}^{2a}+\sum_{a=1}^{2015} \omega_{2}^{2a}+\sum_{a=1}^{2015} 2$ .

Using the formulae for the sum of a geometric progression, we found that it is equal to

$(\frac{\omega_{1}^{2*2016}-1}{\omega_{1}^{2}-1}-1)+(\frac{\omega_{2}^{2*2016}-1}{\omega_{2}^{2}-1}-1)+2*2015$ .

Now $\omega_{1}^{2*2016}=\omega_{1}^{2*3*672}=(\omega_{1}^{3})^{2*672}=(-1)^{2*672}=1$ .

In the same way $\omega_{2}^{2*2016}=1$ .

Thus the second term simplifies to $(0-1)+(0-1)+2*2015=4028$ .

Finally $-3+4028=4025$ .