đặt \(A=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

Ta có :

\(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

Lại có :

\(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)

Tu lam di

Ta có :

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}< 1\)

\(\Rightarrow A< 1\)

ta có:

A=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+....+\(\frac{1}{100^2}\)< B=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+.....+\(\frac{1}{99.100}\)

B=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+........+\(\frac{1}{99}\)-1/100

B=1-1/100=99/100<1

Vì a<b mà B lại bé hơn 1 =>A<1

Do 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 1/1.2 + 1/2.3 + 1/3.4 +...+ 1/99.100 (1)

Mà 1/2 + 1/3 + 1/4 +...+ 1/100= 1-1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/99 - 1/100

= 1-1/100 = 99/100 <1  (2)

Từ 1 và 2 => 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 1

hay A < 1

giải tương tự như câu hôm qua mình giải

để chứng minh A < \(\frac{1}{10}\). Ta thấy \(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)

\(\Rightarrow A^2< \left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)

\(=\frac{1.\left(3.5...99\right)}{2.4.6...100}.\frac{2.4.6...100}{\left(3.5.7...99\right).101}\)

\(=\frac{1}{101}< \frac{1}{10}\)

\(\Rightarrow A^2< \frac{1}{101}< \frac{1}{100}=\frac{1}{10^2}\Rightarrow A< \frac{1}{10}\)

để chứng minh A > \(\frac{1}{15}\). Ta thấy \(A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)

\(\Rightarrow A^2>\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\right)\)

\(=\frac{1.\left(3.5...99\right)}{\left(2.4.6...98\right).100}.\frac{1.\left(2.4...98\right)}{2.\left(3.5...99\right)}\)

\(=\frac{1}{100}.\frac{1}{2}=\frac{1}{200}\)

\(\Rightarrow A^2>\frac{1}{200}>\frac{1}{225}=\frac{1}{15^2}\Rightarrow A>\frac{1}{15}\)

BƯỚC 1   4A

BƯỚC 2   4A-A

BƯỚC  3   3A=1-1/100^2

BƯỚC 4    A =  3-3/100^2

BƯỚC 5   HẢI AI KHÁC Ý

Đặt \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...\frac{1}{100^2}\)

Ta có : 

\(A< \frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+...+\frac{1}{99\times100}\)

\(\Rightarrow A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

Ta có : 

\(A>\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{100\times101}\)

\(\Leftrightarrow A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{100}>\frac{1}{6}\)

Vậy \(\frac{1}{6}< A< \frac{1}{4}\left(đpcm\right)\)

ta có: \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

mà \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

                                                       \(=1-\frac{1}{50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1\)

\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1=2\)

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

\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\left(đpcm\right)\)

CMR là gì vậy chị nếu em biết được thì có thể giải giùm chị em có công thức đây(lớp 5)

CÂU HỎI TƯƠNG TỰ NHA BẠN

Dat A=1/5^2+1/6^2+1/7^2+............1/100^2<1/4.5+1/5.6+1/6.7+....+1/99.10=

1/4-1/5+1/5-1/6+1/6-1/7+.............1/99-1/100=

14-1/100=25/100-1/100=24/25/100=1/4(1)

A>1/5.6+1/6.7+1/7.8+....+1/100.101=

1/5-1/6+1/6-1/7+1/7-1/8 +...+1/100-1/101=

1/5-1/101>6 (2)

Tu 1 va 2 => dieu can chung minh