Let $c_i=2^ia_n+2^{i-1}a_{n-1}+\ldots+a_{n-i}$. We know that $c_{-1}=c_n=0$ and that $c_{i+1}-2c_i=a_{n-i-1}\geq -1$. Suppose that $i$ is the greatest index in $\{0, 1, \ldots n\}$ for which $c_i>0$; but then $c_{i+1}\geq2c_i-1>0$ also. This is a contradiction unless $i=n$, impossible because $c_n=0$. Thus $c_i\leq 0$ for all $i$. But we can't have $c_i=0$ for all $i$, because that happens iff $a_i=0$ for all $i$, which is prohibited. Thus $c_0+c_1+c_2+\ldots+c_{n-1}<0$. Now we note that $0>c_0+c_1+\ldots+c_{n-1}=a_n(2^{n-1}+2^{n-2}+\ldots+1)+$ $a_{n-1}(2^{n-2}+2^{n-3}+\ldots+1)+\ldots+a_1$ $=2^na_n+2^{n-1}a_{n-1}+\ldots+2a_1+a_0-(a_0+a_1+\ldots+a_n)=-(a_0+a_1+\ldots+a_n)$. Hence we're done.