Đặt `B=1/5+1/5^{2}+1/5^{3}+...+1/5^{101}`

`<=>5B=1+1/5+1/5^{2}+...+1/5^{100}`

`<=>5B-B=(1+1/5+1/5^{2}+...+1/5^{100})-(1/5+1/5^{2}+...+1/5^{101})`

`<=>5B-B=1+1/5+1/5^{2}+...+1/5^{100}-1/5-1/5^{2}-...-1/5^{101}`

`<=>4B=1-1/5^{101}`

`<=>B=(1-1/5^{101})/4`

`@Shả`

\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{101}}\)

\(5A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{100}}\)

\(5A-A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{100}}-\dfrac{1}{5}-\dfrac{1}{5^2}-...-\dfrac{1}{5^{101}}=1-\dfrac{1}{5^{101}}\Rightarrow A=\dfrac{1-\dfrac{1}{5^{101}}}{4}\)

Đặt biểu thức trong ngoặc đơn là B

\(5B=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{98}}+\dfrac{1}{5^{99}}\)

\(\Rightarrow4B=5B-B=1-\dfrac{1}{5^{100}}\Rightarrow B=\dfrac{1}{4}\left(1-\dfrac{1}{5^{100}}\right)\)

\(\Rightarrow A=4.5^{100}.\dfrac{1}{4}\left(\dfrac{5^{100}-1}{5^{100}}\right)+1=\)

\(=5^{100}\)

a) Đặt \(C=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{100}}\)

\(\Rightarrow5C=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{99}}\)

\(\Rightarrow5C-C=1-\dfrac{1}{5^{100}}\Rightarrow4C=1-\dfrac{1}{5^{100}}\Rightarrow C=\dfrac{1-\dfrac{1}{5^{100}}}{4}\)

\(\Rightarrow A=8.5^{100}.\dfrac{1-\dfrac{1}{5^{100}}}{4}+1=2.\left(5^{100}-1\right)+1=2.5^{100}-2+1=2.5^{100}-1\)

b)\(B=\dfrac{4}{3}-\dfrac{4}{3^2}+...-\dfrac{4}{3^{100}}\)

\(B=4.\left(\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\right)\)

Đặt \(\left(\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\right)=D\)

\(\Rightarrow3D=1-\dfrac{1}{3}+...-\dfrac{1}{3^{99}}\)

\(\Rightarrow3D+D=1-\dfrac{1}{3^{100}}\)

\(\Rightarrow D=\dfrac{1-\dfrac{1}{3^{100}}}{4}\)

a) `1/3 - 1/4 : 2/5 = 1/3 - 5/8 = -7/24`

b) `6/7-(5/6+1/3)-(2/3+1/7) = 6/7-5/6-1/3-2/3-1/7`

`=(6/7-1/7)-(1/3+2/3)-5/6`

`=5/7-1-5/6`

`=-47/42`

c) `-5/9 . 2/5 + 4 5/9 + 5/9 . (-3/5)`

`= -5/9 . 2/5 + 4 + 5/9 + (-5/9) . 3/5`

`=-5/9 . (2/5 + 3/5-1) + 4`

`=-5/9 . 0 +4`

`=4`

d) 3 1/2 - (5 4/7 - 1 1/2) : 0,75`

`=7/2 - (39/7 - 3/2) : 3/4`

`= 7/2 - 57/14 : 3/4`

`=7/2 - 38/7`

`=-27/14`

De tke nma khonq bt lamm

Nguu

\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{100\cdot101}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}-\dfrac{1}{20}=\dfrac{1}{5}\left(1\right)\)

\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}< \dfrac{1}{3}\)