The given series terms can be described as follows:
[tex]\begin{gathered} A_1=1 \\ A_2=\frac{3}{2} \\ A_3=\frac{7}{4} \\ A_4=\frac{15}{8} \\ \ldots \\ A_n=\frac{2^n-1}{2^{n-1}} \end{gathered}[/tex]From this, we are able to calculate and sum the rest of the terms until A10, as follows:
[tex]\begin{gathered} A_5=\frac{31}{16} \\ A_6=\frac{63}{32} \\ A_7=\frac{127}{64} \\ A_8=\frac{255}{128} \\ A_9=\frac{511}{256} \\ A_{10}=\frac{1023}{512} \\ \\ S=\sum ^{10}_{n\mathop=1}A_n=\frac{9217}{512} \end{gathered}[/tex]From the solution de developed above, we are able to conclude that the answer to the present question is:
[tex]S=\frac{9217}{512}[/tex]