a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

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

Giúp mik vs ạ.Mik đag cần

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

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

a>

\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000

ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )

1/100^2<1/2

=>A<1

Ta thấy :

\(\dfrac{1}{4^2}>\dfrac{1}{4.5}\)

\(\dfrac{1}{5^2}>\dfrac{1}{5.6}\)

..............

\(\dfrac{1}{99^2}>\dfrac{1}{99.100}\)

\(\Rightarrow\) \(K>\dfrac{1}{4.5}+\dfrac{1}{5.6}+.....+\dfrac{1}{99.100}\)

Ta có công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)

Dựa vào công thức ta có :

\(\dfrac{1}{4.5}=\dfrac{1}{4}-\dfrac{1}{5}\)

\(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)

.......................

\(\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow\) \(K>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+......+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Leftrightarrow\) \(K>\dfrac{1}{4}-\dfrac{1}{100}\)

\(\Rightarrow K>\dfrac{6}{25}>\dfrac{1}{5}\Rightarrow dpcm\) (1)

Ta có :

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)

................

\(\dfrac{1}{99^2}< \dfrac{1}{98.99}\)

Dựa vào công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\) ta có :

\(K< \dfrac{1}{3.4}+\dfrac{1}{4.5}+......+\dfrac{1}{98.99}\)

\(\Rightarrow\) \(K< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+.......+\dfrac{1}{98}-\dfrac{1}{99}\)

\(\Rightarrow\) \(K< \dfrac{1}{3}-\dfrac{1}{99}\)

Vậy \(K< \dfrac{32}{99}< \dfrac{1}{3}\Rightarrow dpcm\) (2)

(1) ; (2) \(\Rightarrow\) \(\dfrac{1}{5}< K< \dfrac{1}{3}\)

Ai thấy đúng thì ủng hộ nha !!!

Công thức tổng quát: \(\dfrac{1}{n\left(n+1\right)}< \dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)

=>\(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}< K< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{98.99}\)

=>\(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}< K< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

=>\(\dfrac{1}{4}-\dfrac{1}{100}< K< \dfrac{1}{3}-\dfrac{1}{100}\)

=>\(\dfrac{1}{4}< K< \dfrac{1}{3}\)

=>\(\dfrac{1}{5}< K< \dfrac{1}{3}\left(do\dfrac{1}{4}>\dfrac{1}{5}\right)\)