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30 tháng 4 2018

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

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

Mà A=1+B=>A=1+B<1+1=2

30 tháng 4 2018

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

\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

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

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

vậy \(A=\frac{99}{100}< 2\left(đpcm\right)\)

B)

ta có : \(1=1\)

\(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)

\(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{7}< \frac{1}{4}+...+\frac{1}{4}=\frac{4}{4}=1\)

\(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}< \frac{1}{8}+...+\frac{1}{8}=\frac{8}{8}=1\)

\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{63}< 1\)

tất cả công lại \(\Rightarrow B< 6\)

1 tháng 5 2019

a) Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2014^2}\)

    Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\)

               \(\frac{1}{3^2}< \frac{1}{2.3}\)

                .................

             \(\frac{1}{2014^2}< \frac{1}{2013.2014}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2013.2014}\)

\(\Rightarrow B< 1-\frac{1}{2014}< 1\)

\(\Rightarrow B< 1\)

\(\Rightarrow1+B< 1+1\)

Hay \(A< 2\)

1 tháng 5 2019

C) Ta có: \(\frac{1}{2}< \frac{2}{3}\)

                 \(\frac{3}{4}< \frac{4}{5}\)

                .................

            \(\frac{9999}{10000}< \frac{10000}{10001}\)

\(\Rightarrow C< \frac{2}{3}.\frac{4}{5}.....\frac{10000}{10001}\)

\(\Rightarrow C^2< \left(\frac{1}{2}.\frac{3}{4}.....\frac{9999}{10000}\right).\left(\frac{2}{3}.\frac{4}{5}.....\frac{10000}{10001}\right)\)

\(\Rightarrow C^2< \frac{1}{10001}< \frac{1}{10000}\)

\(\Rightarrow C^2< \frac{1}{10000}\)

\(\Rightarrow C< \frac{1}{100}\)

1 tháng 5 2018

2A=1+1/2+1/2^2+1/2^3+...+1/2^99

-A=    1/2+1/2^2+1/2^3+...+1/2^99+1/2^100

-------------------------------------------------------------------

A=1-1/2^100

A=2^100-1/2^100<1(dpcm)

1 tháng 5 2018

B), B=2/1.2 +22.3 +23.4 +...+299.100 <2 =

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

=1-1/100

=99/100 

vì 99/100<2 nên B=2/1.2+2/2.3+2/3.4+......+2/99.100<2

30 tháng 4 2019

Bài 1 :

\(x\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\right)=1\)

\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)

\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{50}\right)=1\)

\(\Rightarrow x\cdot\frac{24}{50}=1\)

\(\Rightarrow x=1\div\frac{24}{50}=\frac{25}{12}\)

                            #Louis

30 tháng 4 2019

\(\frac{1}{2.3}x+\frac{1}{3.4}x+\frac{1}{4.5}x+...+\frac{1}{49.50}x=1\)

\(\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)x=1\)

\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)x=1\)

\(\left(\frac{1}{2}-\frac{1}{50}\right)x=1\)

\(\frac{12}{25}x=1\)

Đến đây dễ rồi :)))

Bn tự tính típ nha

1 tháng 5 2018

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

\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

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

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

\(A=\frac{99}{100}< 2\left(đpcm\right)\)

2 tháng 5 2018

Ta có:

\(C=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\)

Đặt \(I=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\)

Ta có: \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};.....;\frac{9999}{10000}< \frac{10000}{10001}\)

\(\Rightarrow C< D\)

Lại có: \(C\cdot D=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\right)\cdot\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\right)\)

\(\Leftrightarrow C\cdot D=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\)

\(\Leftrightarrow C\cdot D=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\cdot\frac{10000}{10001}\)

\(\Leftrightarrow C\cdot D=\frac{1}{10001}\)

Mà C<D \(\Rightarrow C\cdot C< C\cdot D\)

Hay \(C\cdot C< \frac{1}{10001}\)

\(\Rightarrow C< \frac{1}{10001}< \frac{1}{100}\)

Vậy \(C< \frac{1}{100}\left(đpcm\right)\)

22 tháng 6 2017

Đặt :\(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)

\(N=\frac{2}{3}.\frac{4}{5}...\frac{10000}{10001}\)

Ta thấy:\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};....;\frac{9999}{10000}< \frac{10000}{10001}\)

Mặt khác ta thấy:

\(C.N=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{10000}{10001}\right)\)

\(C.N=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{9999}{10000}.\frac{10000}{10001}\)

\(C.N=\frac{1.2.3....9999.10000}{2.3.4....10000.10001}\)

Rút gọn  phép tính \(C.N\)

\(C.N=\frac{1}{10001}\)

\(C.C< N\Rightarrow C.C< C.N\)

Hay\(C.C< \frac{1}{10001}< \frac{1}{10000}=\frac{1}{10}.\frac{1}{10}\)

\(\Rightarrow C< \frac{1}{10000}\)(đpcm)