A) 52 . B) 56 . C) 54. D)58 . E)None the these
Where is this concern from? I"m pretty sure it originates from one high institution centregalilee.com contest, go anyone one know which centregalilee.com contest and of course i can"t solve, I have an answer key but ns don"t understand the solution.
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Multiply both political parties by $x-1\ne 0$:$$x^2+x+1=0 \Rightarrow (x-1)(x^2+x+1)=0 \Rightarrow x^3-1=0 \Rightarrow x^3=1\Rightarrow \\x^54=1;x^-54=1 \quad (1)$$Expand:$$(x+\frac1x)^2+(x^2+\frac1x^2)^2+(x^3+\frac1x^3)^2+\ldots+(x^27+\frac1x^27)^2=\\
$x^2+1=-x\Rightarrow x+\frac1x=-1\Rightarrow \frac1x=-x-1=x^2$ $<$as $x \ne 0>$.
$x^n+\frac1x^n=-1$ together $n = 3k+1,3k+2$
$x^n+\frac1x^n=2$ together $n = 3k$
So, below the price is $18+4 \times9=54$
We need $\sum_r=0^8\sum_n=1^3(x^3r+n+x^-(3r+n))^2$
Too lengthy for a comment.
Since you currently received great explanations, i just concentrated on the an ext general problem of$$S_n=\sum_k=1^n \left(x^k+x^-k\right)^2 \qquad \textwhere\qquad x^2+x+1=0\implies x=i^4/3$$These room the number which are congruent to $\0, 1, 2\$ modulo $6$ and there room several methods to write them. If girlfriend look here, you will find the nice$$S_n= 2 n-1+\frac1\sqrt3\sin \left(\frac2 \pi 3n\right)+\cos \left(\frac2 \pi 3n\right)$$
$$S_n=n-1+3 \left\lfloor \fracn-13\right\rfloor $$
edited january 23 "19 in ~ 3:54
answered jan 22 "19 in ~ 11:39
Claude LeiboviciClaude Leibovici
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Thank you because that all her answers, ns just discover the other solution to deal with this, for this reason I desire to article it here.
Since $x^2+x+1=0$, so we can uncover
and $x^2+\frac1x^2=(x+\frac1x)^2-2=(-1)^2-2=-1$. Let"s save going
So it has a period,-1,-1,2,-1,-1,2,-1,-1,-2.......until the critical term.
The sum of the one duration is $(-1)^2+(-1)^2+(2)^2=6$, and also we have actually 27$\div$ 3=9, so the final value is equal to 6$\times$9=54.
edited Oct 8 "19 in ~ 17:19
answered Oct 8 "19 at 12:01
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