Tính `A=1+2/2^2+3/2^3+4/2^4+…+100/2^100`. 02/08/2021 Bởi Madelyn Tính `A=1+2/2^2+3/2^3+4/2^4+…+100/2^100`.
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 $A=\dfrac{5}{2}-\dfrac{1}{2^ 99}-\dfrac{100}{2^100}$ Bình luận
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 $A=\dfrac{5}{2}-\dfrac{1}{2^ 99}-\dfrac{100}{2^100}$