Interesting summation question: If $a$ and $b$ are the roots of $x^2+x+1$, then what is the below expression equal to?

Question

Question: If $a$ and $b$ are the roots of $x^2+x+1$, then what is the below expression equal to? $$\sum_{n=1}^{1729} \left[(-1)^n\cdot V(n)\right]$$ Where $$V(n)=a^n+b^n$$

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My effort:

I think I managed to solve this, but in an inefficient way. I found $a$ and $b$, calculated $V(1)$,$V(2)$, and so on; after that, I calculated $S_1$,$S_2$ and so on, where $S_n=V(1)+V(2) \cdot\cdot\cdot +V(n)$

I found that

$S_1$,$S_7$,$S_{13}$ $\cdot\cdot\cdot$ resulted in $1$;

$S_2$,$S_8$,$S_{14}$ $\cdot\cdot\cdot$ resulted in $0$;

$S_3$,$S_9$,$S_{15}$ $\cdot\cdot\cdot$ resulted in $-2$;

$S_4$,$S_{10}$,$S_{16}$ $\cdot\cdot\cdot$ resulted in $-3$;

$S_5$,$S_{11}$,$S_{17}$ $\cdot\cdot\cdot$ resulted in $-2$;

$S_6$,$S_{12}$,$S_{18}$ $\cdot\cdot\cdot$ resulted in $0$;

Since $S_{1729}$ would come under the first sequence, I think the answer to the question is $1$.

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My request: Could someone please suggest a more efficient method to tackle/solve this problem?

Thank you in advance.

Answer 1

HINT:

$x^2+x+1=0\implies x^2=-x-1,x^3=x(-x-1)=-x^2-x=1\implies a^3=b^3=1$

Using Exponent Combination Laws/Power of Product, $(-1)^na^n=(-a)^n,$

$\displaystyle\implies\sum_{n=1}^{1729} \left[(-1)^n\cdot V(n)\right]=\sum_{n=1}^{1729}[(-a)^n+(-b)^n]$

As $\displaystyle1730=3\cdot576+2,a^{1730}=\cdots a^2$

$\displaystyle\sum_{n=1}^{1729}(-a)^n=\dfrac{1-(-a)^{1730}}{1-(-a)}=\cdots=1-a$ as $1+a\ne0$

Hope you can take it home here!

Answer 2

Since $a$ is a root of $x^2+x+1$, we have $a^2+a+1 = 0$.

Hence, $a^3-1 = (a-1)(a^2+a+1) = 0$, and thus, $a^{n+3}-a^n = a^n(a^3-1) = 0$.

Therefore, $a^{n+3} = a^n$. Similarly, $b^{n+3} = b^n$. Thus, $V(n+3) = V(n)$ for all integers $n$.

Now, write the sum as $\displaystyle\sum_{n = 1}^{1729}(-1)^nV(n) =$ $-V(1) + \displaystyle\sum_{k = 0}^{287}\left[V(6k+2)-V(6k+3)+V(6k+4)-V(6k+5)+V(6k+6)-V(6k+7)\right]$

For each $k$, we have $V(6k+2) = V(6k+5)$, and $V(6k+3) = V(6k+6)$, and $V(6k+4) = V(6k+7)$. So for each integer $k$, we have $V(6k+2)-V(6k+3)+V(6k+4)-V(6k+5)+V(6k+6)-V(6k+7) = 0$.

Thus, the summation is simply $-V(1)$.

Answer 3

(Just another possible method...)

Let $f(n) = (-1)^n ( a^n + b^n ) = (-a)^n + (-b)^n$ for any $n \in \mathbb{Z}$.

Then $f(n+2) = f(n+1) - f(n)$ because $(-a)$ and $(-b)$ are roots of $r \mapsto r^2 - r + 1$.

Since $f(0) = 1$ and $f(1) = 1$, we get $f$ generating $1,1,0,-1,-1,0,...$ (repeats every $6$ terms).

Also, $\sum_{n=1}^k f(n) = \sum_{n=1}^k ( f(n+1) - f(n+2) ) = f(2) - f(k+2) = -f(k+2)$.

For $k = 1729$, we get $\sum_{n=1}^k f(n) = -f(3) = 1$.

Answer 4

Another way to look at the problem could be the following.

The roots of $x^2+x+1=0$ are given by $$a=-\frac{1}{2}-\frac{i \sqrt{3}}{2}=e^{-\frac{2 i \pi }{3}}$$ $$b=-\frac{1}{2}+\frac{i \sqrt{3}}{2}=e^{+\frac{2 i \pi }{3}}$$ So $$a^n=e^{-\frac{2 n i \pi }{3}}$$ $$b^n=e^{+\frac{2 n i \pi }{3}}$$ $$a^n+b^n=2 \cos \left(\frac{2 \pi n}{3}\right)$$ $$S_p=\sum_{n=1}^p (-1)^n (a^n+b^n)=2 \cos \left(\frac{(5 p+1)\pi}{3} \right)-1$$