smc

For every sequence $S=\left(a_1,a_2,\dots ,a_n\right)$ of at least three terms, $$A^2(S)=\left(\frac{a_1+2a_2+a_3}{4},\frac{a_2+2a_3+a_4}{4},\dots ,\frac{a_{n-2}+2a_{n-1}+a_n}{4}\right).$$Thus for $m=1\text{ and }2$, the coefficients of the terms in the numerator of $A^m(S)$ are the binomial coefficients $\left(\genfrac{}{}{0ex}{}{m}{0}\right),\left(\genfrac{}{}{0ex}{}{m}{1}\right),\dots ,\left(\genfrac{}{}{0ex}{}{m}{m}\right)$, and the denominator is $2^m$. Because $\left(\genfrac{}{}{0ex}{}{m}{r}\right)+\left(\genfrac{}{}{0ex}{}{m}{r+1}\right)=\left(\genfrac{}{}{0ex}{}{m+1}{r+1}\right)$ for all integers $r\ge 0$, the coefficients of the terms in the numerators of $A^{m+1}(S)$ are $\left(\genfrac{}{}{0ex}{}{m+1}{0}\right),\left(\genfrac{}{}{0ex}{}{m+1}{1}\right),\dots ,\left(\genfrac{}{}{0ex}{}{m+1}{m+1}\right)$ for $2\le m\le n-2$. The definition implies that the denominator of each term in $A^{m+1}(S)$ is $2^{m+1}$. For the given sequence, the sole term in $A^{100}(S)$ is $$\frac{1}{2^{100}}\sum _{m=0}^{100}\left(\genfrac{}{}{0ex}{}{100}{m}\right)a_{m+1}=\frac{1}{2^{100}}\sum _{m=0}^{100}\left(\genfrac{}{}{0ex}{}{100}{m}\right)x^m=\frac{1}{2^{100}}(x+1{)}^{100}.$$Therefore, $$\left(\frac{1}{2^{50}}\right)=A^{100}(S)=\left(\frac{(1+x{)}^{100}}{2^{100}}\right),$$so $(1+x{)}^{100}=2^{50}$, and because $x>0$, we have $x=\boxed{\sqrt{2}-1}$.