Đáp án:

Vậy $A=\dfrac{5}{2}-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}$

Giải thích các bước giải:

Ta có: $A=1+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{100}{2^100}$

$=>2A=2.(1+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{100}{2^100})$

$=>2A=2.1+2.\dfrac{2}{2^2}+2.\dfrac{3}{2^3}+2.\dfrac{4}{2^4}+...+2.\dfrac{100}{2^100}$

$=>2A=2+\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{100}{2^99}$

$=>2A-A=(2+\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{100}{2^99})-(1+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{100}{2^100})$

$=>(2-1).A=2+\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{100}{2^99}-1-\dfrac{2}{2^2}-\dfrac{3}{2^3}-\dfrac{4}{2^4}-...-\dfrac{100}{2^100}$

$=>A=(2-1)+\dfrac{2}{2}+(\dfrac{3}{2^2}-\dfrac{2}{2^2})+(\dfrac{4}{2^3}-\dfrac{3}{2^3})+...+(\dfrac{100}{2^99}-\dfrac{99}{2^99})-\dfrac{100}{2^100}$

$=>A=1+\dfrac{2}{2}+\dfrac{1}{2^2})+\dfrac{1}{2^3}+...+\dfrac{1}{2^99}-\dfrac{100}{2^100}$

Đặt $B=\dfrac{1}{2^2})+\dfrac{1}{2^3})+...+\dfrac{1}{2^99}$

$=>\dfrac{1}{2}B=\dfrac{1}{2}.\dfrac{1}{2^2})+\dfrac{1}{2^3}+...+\dfrac{1}{2^99}$

$=>\dfrac{1}{2}B=\dfrac{1}{2^3})+\dfrac{1}{2^4})+...+\dfrac{1}{2^100}$

$=>B-\dfrac{1}{2}B=(\dfrac{1}{2^2})+\dfrac{1}{2^3})+...+\dfrac{1}{2^99})-(\dfrac{1}{2^3})+\dfrac{1}{2^4})+...+\dfrac{1}{2^100})$

$=>\dfrac{1}{2}B=\dfrac{1}{2^2})-\dfrac{1}{2^100}$

$=>B=2.(\dfrac{1}{2^2})-\dfrac{1}{2^100})$

$=>B=\dfrac{1}{2}-\dfrac{1}{2^99}$

Quay lại với A:

$=>A=1+\dfrac{2}{2}+(\dfrac{1}{2}-\dfrac{1}{2^99})-\dfrac{100}{2^100}$

$=>A=2+\dfrac{1}{2}-\dfrac{1}{2^99}-\dfrac{100}{2^100}$

$=>A=\dfrac{5}{2}-\dfrac{1}{2^99}-\dfrac{100}{2^100}$

Vậy $A=\dfrac{5}{2}-\dfrac{1}{2^ 99}-\dfrac{100}{2^100}$