Bài giải:

D = $\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+...+\frac{1025}{1024}$

Vì số: 1024 = 2 x 2 x 2 x 2 x … x 2 (có 10 số 2 nhân với nhau) nên tổng trên có 10 số hạng.

Ta thấy: $\frac{3}{2}=1+\frac{1}{2}$ ; $\frac{5}{4}=1+\frac{1}{4}$ ; $\frac{9}{8}=1+\frac{1}{8}$ ; … ; $\frac{1025}{1024}=1+\frac{1}{1024}$

D = $\left( 1+\frac{1}{2} \right)+\left( 1+\frac{1}{4} \right)+\left( 1+\frac{1}{8} \right)+\left( 1+\frac{1}{16} \right)+...+\left( 1+\frac{1}{1024} \right)$

D = $(1+1+1+1+...+1)+\left( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{1024} \right)$

D = 10 + $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{1024}$

Đặt S = $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{1024}$

2 x S = $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}$

2 x S – S = $\left( 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512} \right)-\left( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{1024} \right)$

S = $1-\frac{1}{1024}=\frac{1023}{1024}$

Vậy D = 10 + $\frac{1023}{1024}=\frac{11263}{1024}$