Sum it up to 1729

$a=\omega \quad \& \quad b={ \omega }^{ 2 }\quad ,where\\ \\ \omega \quad \& \quad \quad { \omega }^{ 2 }\quad are\quad complex\quad root\quad of\quad { x }^{ 3 }=1\\ \\ { \left( -1 \right) }^{ n }{ V }_{ n }={ \left( -\omega \right) }^{ n }{ +\left( { -\omega }^{ 2 } \right) }^{ n }\\ \\ \Rightarrow \sum _{ n=0 }^{ 1729 }{ { \left( -1 \right) }^{ n }{ V }_{ n }= } \sum _{ n=0 }^{ 1729 }{ { \left( -\omega \right) }^{ n } } +\sum _{ n=0 }^{ 1729 }{ { \left( { -\omega }^{ 2 } \right) }^{ n } } \\ \\ \quad \quad \quad =\cfrac { { 1-\left( -\omega \right) }^{ 1730 } }{ 1-\left( -\omega \right) } +\cfrac { { 1-\left( -{ \omega }^{ 2 } \right) }^{ 1730 } }{ 1-\left( -{ \omega }^{ 2 } \right) } \\ \\ \quad \quad =\cfrac { 1-{ \omega }^{ 1730 } }{ 1+\omega } +\cfrac { 1-{ \omega }^{ 3460 } }{ 1+{ \omega }^{ 2 } } \\ \\ \quad \quad =\cfrac { 1-{ \omega }^{ 2 } }{ 1+\omega } +\cfrac { 1-\omega }{ 1+{ \omega }^{ 2 } } \quad \left( as\quad { \omega }^{ 3 }=1 \right) \\ \\ \quad =\cfrac { 1-{ \omega }^{ 4 }+1-{ \omega }^{ 2 } }{ 1+\omega +{ \omega }^{ 2 }+{ \omega }^{ 3 } } \\ \\ \quad =\cfrac { 2-{ \omega }^{ 2 }-\omega }{ 0+1` } \quad \left( as\quad 1+\omega +{ \omega }^{ 2 }=0\quad \& \quad { \omega }^{ 3 }=1 \right) \\ \\ \quad =3\quad \left( as\quad 1+\omega +{ \omega }^{ 2 }=0\quad \right) \\ \\ \quad$

The roots of $x^2+x+1 = 0$ , $\quad (a,b) = - \frac {1}{2} \pm i \frac {\sqrt{3}}{2}$ , where $i = \sqrt{-1}$ .

Using the Euler's identity, we have: $(a,b) = - \frac {1}{2} \pm i \frac {\sqrt{3}}{2} = \cos {\frac {2\pi}{3}} \pm \sin {\frac {2\pi}{3}} = e^{\pm i \frac{2\pi}{3}}$ .

$\Rightarrow V_n = a^n + b^n = e^{i \frac{2n\pi}{3}} + e^{-i \frac{2n\pi}{3}} = 2\cos {\frac{2n\pi}{3}}$

$\Rightarrow V_n = \begin{cases} +2 \quad & n \equiv 0 \pmod {3} \\ -1 \quad & n \equiv 1 \pmod {3} \\ -1 \quad & n \equiv 2 \pmod {3} \end{cases}$

Similarly,

$\sum _{n-0} ^N {(-1)^n V_n} = \begin{cases} 2 \quad & N \equiv 0 \pmod {6} \\ 3 \quad & N \equiv 1 \pmod {6} \\ 2 \quad & N \equiv 2 \pmod {6} \\ 0 \quad & N \equiv 3 \pmod {6} \\ -1 \quad & N \equiv 4 \pmod {6} \\ 0 \quad & N \equiv 5 \pmod {6} \end{cases}$

Since $1729 \equiv 1 \pmod {6}$ , the answer is $\boxed {3}$ .

Solution without using complex numbers : First note that $V_{n+1}=a^{n+1}+b^{n+1}=(a+b)(a^n+b^n)-ab(a^{n-1}+b^{n-1})=-V_{n}-V_{n-1}$ i.e., $V_{n+1}+V_n+V_{n-1}=0$ Writing the above equality for $n\gets n+1$ , we have $V_{n+2}+V_{n+1}+V_n=0$ Comparing the above two equations, we have $V_{n-1}=V_{n+2}$ , i.e. $V_n=V_{n+3}$ Next we form groups of six consecutive terms in the summation $\sum_{n=0}^{1729}(-1)^nV_n$ and utilize the above observation to realize that the sum of each group is zero, e.g. $V_0-V_1+V_2-V_3+V_4-V_5=0.$ Hence the value of the given sum is last two left-over terms, viz $\sum_{n=0}^{1729}(-1)^nV_n=V_{1728}-V_{1729}=V_0-V_1=2-(-1)=3$ .

I used de moivre's theorem !

Here is a solution using Newton's identities. One form of the general identities is: $p_k = \sum_{i=k-n}^{k-1} (-1)^{k-1+i} e_{k-i} p_i$ where $p_i(a,b) = a^i + b^i$ and $e_i$ is the ith elementary symmetric polynomial. This gives the following sequence of equations: $p_1 = e_1 \\ p_2 = e_1 p_1 - 2 e_2 \\ p_3 = e_1 p_2 - e_2 p_1 + 3 e_3 \\ \vdots$ Since we are dealing with the roots of a quadratic, by definition $e_i = 0$ for $i > 2$ . This then simplifies to: $p_k = e_1 p_{k-1} - e_2 p_{k-2}$ for $k \ge 3$ . This gives us a recurrence for calculating values of $p_k$ , but we first need to determine the values of $e_1$ and $e_2$ . Since they are symmetric polynomials of the roots of a polynomial we can use Vieta's formulas. Since $a, b$ are the roots of $x^2 + x + 1$ we have $e_1 = p_1 = -1$ and $e_2 = 1$ . This gives us a complete recurrence for $p_k$ : $p_k = -p_{k-1} - p_{k-2}, k \ge 2 \\ p_0 = 2, p_1 = -1$ This is a second order homogeneous linear recurrence relation with constant coefficients and so a general solution can be found, but for the purposes of this problem it is enough to note that $p_3 = p_0, p_4 = p_1, p_5 = p_2 = -1$ and since each term relies only on the previous two, therefore the sequence must be periodic with period 3. That is, the sequence of $p_k$ satisfies: $p_k = \begin{cases} 2 & k \equiv 0 \pmod 3 \\ -1 & \mathrm{otherwise} \end{cases}$ Clearly, from our definition $p_i = V_i$ , so the value we are interested in is: $\sum_{n=0}^{1729} (-1)^n p_n$ Since this sequence has period 3 (and hence also has period 6), $\sum_{n=a}^{a+5} (-1)^n p_n = 0$ for any $a$ , and $1729 \equiv 1 \pmod 6$ , this becomes: $\sum_{n=0}^{1729} (-1)^n p_n = 288 \sum_{n=0}^5 (-1)^n p_n + p^{1728} - p^{1729} \\ = 0 + 2 + 1 = 3$ Therefore, the answer is $\boxed{3}$ .

Let $N:=1730$ as a short-hand. You can do most of the job using the geometric sum:

$\begin{aligned} S:=\sum_{n=0}^{N-1}(-1)^nV_n&=\sum_{n=0}^{N-1}(-a)^n+(-b)^n=\frac{1-(-a)^N}{1+a}+\frac{1-(-b)^N}{1+b}=\frac{1-(-a)^{R}}{1+a}+\frac{1-(-b)^R}{1+b},\qquad N\equiv R\mod 6 \end{aligned}$

Use $a^3=b^3=1$ in the last simplification. Now observe $N=1730\equiv 2\mod 6$ to obtain

$\begin{aligned} S&=\frac{1-a^2}{1+a}+\frac{1-b^2}{1+b}\underset{\text{3. Binomi}}{=}1-a+1-b\underset{\text{Vieta}}{=}2+1=\fbox{3} \end{aligned}$

Rem.: You can use the first equation together with Vieta's formula to gain an expression for any $N\in\mathbb{N}$ :

$\begin{aligned} S=\begin{cases} 0,&N\in\{0;\:4\}\mod 6\\ 2,&N\in\{1;\:3\}\mod 6\\ 3,&N\equiv 2\mod 6\\ -1,&N\equiv 5\mod 6 \end{cases} \end{aligned}$

note that vn always equal to -1

v0=2