Đáp án:

$S<2$.

Giải thích các bước giải:

$S=\dfrac12+\dfrac2{2^2}+\dfrac3{2^3}\ +\,.\!.\!.+\ \dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\\\Rightarrow 2S=2.\!\left(\dfrac12+\dfrac2{2^2}+\dfrac3{2^3}\ +\,.\!.\!.+\ \dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\right)\\\Rightarrow 2S=1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\\\Rightarrow 2S-S=\left(1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\right)-\left(\dfrac12+\dfrac2{2^2}+\dfrac3{2^3}\ +\,.\!.\!.+\ \dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\right)\\\Rightarrow S=1+\dfrac12+\dfrac1{2^2}+\dfrac1{2^3}\ +\,.\!.\!.+\ \dfrac1{2^{99}}-\dfrac{100}{2^{100}}\\\Rightarrow 2S=2.\!\left(1+\dfrac12+\dfrac1{2^2}+\dfrac1{2^3}\ +\,.\!.\!.+\ \dfrac1{2^{99}}-\dfrac{100}{2^{100}}\right)\\\Rightarrow 2S=2+1+\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{98}}-\dfrac{100}{2^{99}}\\\Rightarrow 2S-S=\left(2+1+\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{98}}-\dfrac{100}{2^{99}}\right)-\left(1+\dfrac12+\dfrac1{2^2}+\dfrac1{2^3}\ +\,.\!.\!.+\ \dfrac1{2^{99}}-\dfrac{100}{2^{100}}\right)\\\Rightarrow S=2-\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\\\Rightarrow S=2-\left(\dfrac{198}{2^{100}}-\dfrac{100}{2^{100}}\right)\\\Rightarrow S=2-\dfrac{98}{2^{100}}<2\\\Rightarrow S<2$

Đáp án:

`S=1/2+2/2²+3/2³+...+99/2⁹⁹+100/2¹⁰⁰`

`2S=1+2/2+3/2^2+....+99/2^98+100/2^99`

`S=1+1/2+1/2^2+1/2^3+....+1/2^99-100/2^100`

`2S=2+1+1/2+1/2^2+...+1/2^100-100/2^99`

`S=2-100/2^99-99/2^100<2`

Vậy` S<2`

Giải thích các bước giải: