Assume that $v_1, \dots, v_n \in \mathbb{R} \setminus \{0\}$ are distinct real numbers. Define $$ v^k = (v_1^k, \dots, v_n^k) \in \mathbb{R}^n. $$
Assume that $x,y \in \mathbb{R}^n$ are two arbitrary vectors such that $$ \langle x,v^{2j+1} \rangle = 0 = \langle y,v^{2j} \rangle \ \ \ \ \forall j = 0,1,2,\dots $$
That means the scalar product of $x$ with $v^k$ is equal to zero for all odd values of $k$ and the scalar product of $y$ with $v^k$ is equal to zero for all even values of $k$. Can we conclude that $x$ and $y$ are equal to zero?
Let's take $v_1= 1, v_2=-1$ and $v_i$ what you want for $i \geq 3$. In this way you get
Now, consider $y = (1,-1,0, \dots, 0) \in \mathbb{R^n}$, then $\langle y,v^{2j} \rangle = 0$ for every $j$.
Analogously you can take $x = (1,1,0,\dots,0)$ and observe that $\langle x,v^{2j+1} \rangle = 0$ for every $j$.
These should be counterexamples of your statement.
P.S.: doing this I supposed $\langle \cdot, \cdot \rangle $ is the standard scalar product , i.e. $\langle x,y\rangle = \sum_{i=1}^n x_i y_i$.
As Achille suggested you have to require a stronger condition in order to get a true statement.