Ques. -3

$\left (\omega^r+\dfrac{1}{\omega^r}\right )^2=1$ if $r\equiv 1,2\mod 3$ $\left (\omega^r+\dfrac{1}{\omega^r}\right )^2=4$ if $r\equiv 0\mod 3$

Thus, the answer will be $8×4+1×17=49$ , as there are $8$ multiples of $3$ in $[1,25]$

${ x }^{ 2 }+x+1=0\quad then\\ x\quad =\quad \frac { 1\pm i\sqrt { 3 } }{ 2 } =\quad cos60+i\quad sin60\quad ={ e }^{ i60 }\\ \therefore (x+\frac { 1 }{ x } )\quad =\quad 2\quad cos60\quad from\quad \\ De-moivre's\quad theorem\quad ({ x }^{ n }+\frac { 1 }{ { x }^{ n } } )=2\quad cos(60n)\\ \therefore { ({ x }^{ n }+\frac { 1 }{ { x }^{ n } } ) }^{ 2 }=\quad 1\quad if\quad n\quad mod\quad 3\quad \equiv \quad 1\quad or\quad 2,\\ \quad { ({ x }^{ n }+\frac { 1 }{ { x }^{ n } } ) }^{ 2 }=\quad 4\quad if\quad n\quad mod\quad 3\equiv 0.\\ \therefore \sum _{ n=1 }^{ 25 }{ { ({ x }^{ n }+\frac { 1 }{ { x }^{ n } } ) }^{ 2 } } =\quad 4\times 4\quad +\quad 4\times 8\quad +\quad 1\quad =\quad 49.\quad$

Using the Euler identity:

${ e}^{ i \theta }=\cos{\theta}+i\sin\theta$ and ${ e}^{- i \theta }=\cos{\theta}-i\sin\theta$

we find:

${ e}^{ i \theta }+ { e}^{- i \theta }=2\cos{\theta}$

for $x^2+x+1=0 \rightarrow x=\frac{1\pm i\sqrt{3}}{2}={ e}^{\pm i \frac{\pi}{3} }$

Now we simply plug and chug:

$\displaystyle \sum _{ r=1 }^{ 25 } \left({ e}^{ i \frac{\pi}{3}r} +\frac{1}{{ e}^{ i \frac{\pi}{3}r}}\right)^2 =\displaystyle \sum _{ r=1 }^{ 25 } \left({ e}^{ i \frac{\pi}{3}r} +{ e}^{ -i \frac{\pi}{3}r}\right)^2 = \displaystyle \sum _{ r=1 }^{ 25 } \left(2\cos {\frac{\pi}{3}r}\right)^2$

or:

$\displaystyle \sum _{ r=1 }^{ 25 } \left({ e}^{ i \frac{\pi}{3}r} +{ e}^{ -i \frac{\pi}{3}r}\right)^2 = \displaystyle \sum _{ r=1 }^{ 25 } \left({ e}^{ i \frac{2\pi}{3}r} +{ e}^{ -i \frac{2\pi}{3}r} +2\right) = \displaystyle \sum _{ r=1 }^{ 25 } \left( 2\cos {\frac{2\pi}{3}r} + 2\right)$

$\displaystyle \sum _{ r=1 }^{ 25 } \left( 2\cos {\frac{2\pi}{3}r} + 2\right) =2 \left( \displaystyle \sum _{ r=1 }^{ 25 } \cos {\frac{2\pi}{3}r} + 25 \right)$

when r is not a multiple of 3: $\cos {\frac{2\pi}{3}r} = -\frac{1}{2}$ (which is a set of 17 possible numbers)

when r is a multiple of 3: $\cos {\frac{2\pi}{3}r} = 1$ (which is a set of 8 possible numbers)

$\therefore 2 \left( \displaystyle \sum _{ r=1 }^{ 25 } \cos {\frac{2\pi}{3}r} + 25 \right) = 2 \left( 17(\frac{-1}{2})+8(1) + 25 \right) = 49$