Question

[Nguyễn Đức Anh]

15 mik hết giờ rồi giúp mik

Answer 1

[Yen Nhi]

`Answer:`

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)

\(\Rightarrow2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{100}}\)

\(B=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{13}{60}\)

\(\Rightarrow B=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}+\frac{13}{60}\)

\(\Rightarrow B=1\)

Mà \(1-\frac{1}{2^{100}}< 1\)

`=>A<B`

Answer 2

[Nguyễn Hà Trang]

$A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}$

$\Rightarrow 2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)$

$\Rightarrow 2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}$

$\Rightarrow 2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)$

$\Rightarrow A=1-\frac{1}{2^{100}}$

$B=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{13}{60}$

$\Rightarrow B=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}+\frac{13}{60}$

$\Rightarrow B=$

Mà $1-\frac{1}{2^{100}}<1$

$\Rightarrow A<B$