Induction: 1. $ a_{0} + 2a_{1} = 0$ $ \Leftrightarrow (a_0,a_1) = (2, - 1)$. (If $ a_1\ge 0$, then $ a_0$ is at least $ - 1$ but $ \textrm{even} - 1$ can't be $ 0$.) 2. $ a_0 + 2a_1 + 2^2 a_2 + \cdots + 2^n a_n = 0$ $ \Leftrightarrow a_0 + 2(a_1 + 2a_2 + \cdots + 2^{n - 1}a_n) = 0$ Thus $a_0$ is even, so $ a_0 = 2k$ for some integer $ k\geq 0$ and therefore $ (a_1 + k) + 2a_2 + \cdots + 2^{n - 1}a_n = 0$. Obviously, $ a_1 + k \ge - 1$ and then by the induction hypothesis we have $ k + a_1 + a_2 + \cdots + a_n > 0$, but $ 2k \geq k$ $ \Rightarrow a_{0} + a_{1} + \cdots + a_{n} > 0$, too

Also induction. $2| a_0$, so $a_0\geq 0, a_0/2\geq 0$. And we have $(a_0/2 + a_1) + 2a_2 +\ldots + 2^{n-1}a_n = 0$, $a_0/2 + a_1\geq -1$. So by induction hypothesis we get that $a_0/2 + a_1+a_2+\ldots + a_n\geq 0$ and so ($a_0\geq 0$) $a_0+\ldots + a_n\geq 0$. Done

BG Yoda wrote: Induction: 1. $ a_{0} + 2a_{1} = 0$ $ \Leftrightarrow (a_0,a_1) = (2, - 1)$. (If $ a_1\ge 0$, then $ a_0$ is at least $ - 1$ but $ \textrm{even} - 1$ can't be $ 0$.) 2. $ a_0 + 2a_1 + 2^2 a_2 + \cdots + 2^n a_n = 0$ $ \Leftrightarrow a_0 + 2(a_1 + 2a_2 + \cdots + 2^{n - 1}a_n) = 0$ Thus $a_0$ is even, so $ a_0 = 2k$ for some integer $ k\geq 0$ and therefore $ (a_1 + k) + 2a_2 + \cdots + 2^{n - 1}a_n = 0$. Obviously, $ a_1 + k \ge - 1$ and then by the induction hypothesis we have $ k + a_1 + a_2 + \cdots + a_n > 0$, but $ 2k \geq k$ $ \Rightarrow a_{0} + a_{1} + \cdots + a_{n} > 0$, too the same solution i had