C = ( 1 - 1/4 ) + ( 1  - 1/9 ) + ( 1 - 1/16 ) + .. .+ ( 1 - 1/10000 )

C = 1 + 1 + ... + 1 - ( 1/4 + 1/9 + 1/16 + ... + 1/10000 )

C = 1 + 1 + 1 +... + 1 - ( 1/2² + 1/3² + .. + 1/100² )

C = 99 - ( 1/2² + 1/3² + ... + 1/100² ) 

Mà 1/2² + 1/3² + ... + 1/100² < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 = 1 - 1/2 + 1/2 - 1/3 + .. + 1/99 - 1/100 = 1 - 1/100 < 1 =>

C > 99 - 1 => C > 98

Ta có:

\(\dfrac{n^2-1}{n^2}=1-\dfrac{1}{n^2}>1-\dfrac{1}{\left(n-1\right)n}\)

Áp dụng:

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

\(C>1-\dfrac{1}{1.2}+1-\dfrac{1}{2.3}+1-\dfrac{1}{3.4}+...+1-\dfrac{1}{99.100}\)

\(C>99-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(C>99-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(C>99-\left(1-\dfrac{1}{100}\right)\)

\(C>98+\dfrac{1}{100}>98\) (đpcm)

đề đúng rồi

\(C=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)

\(C=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)

\(C=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)

\(C=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)

đặt \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\)

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

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

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

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

\(\Rightarrow A< 1\)

Vì \(A< 1\)nên \(B=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)>99-1=98\)

= 3/2² + 8/3² + 15/4² + ... + 9999/100²

= 1.3/2.2 + 2.4/3.3 + 3.5/4.4 + .... + 99.101/100.100

\(=\frac{1.3.2.4.3.5.4.6...99.101}{2^2.3^2....100^2}=\frac{1.2.3^2.4^2...99^2.100.101}{2^2.3^2....100^2}=\frac{1.2.101}{2^2.100}=\frac{101}{200}\)

Vậy C < 98

Pạn coi lại 

Trả lời

Ta có:

\(C=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)

\(\Rightarrow C=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)

\(\Rightarrow C=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)(99 chữ số 1)

\(\Rightarrow C=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)

Ta lại có:

\(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

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

\(\Rightarrow D< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(\Rightarrow D< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow D< 1-\frac{1}{100}\)

\(\Rightarrow D< \frac{99}{100}< 1\)

\(\Rightarrow C>99-1\)

\(\Rightarrow C>98\)

Vậy C>98 (đpcm)

\(C=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)

Ta có: \(\frac{3}{4}=1-\frac{1}{4};\frac{8}{9}=1-\frac{1}{9};\frac{15}{16}=1-\frac{1}{16};...;\frac{9999}{10000}=1-\frac{1}{10000}\)

=> \(C=1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+1-\frac{1}{10000}\)

=> \(C=\left(1+1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)(99 chữ số 1)

=> \(C=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)

Ta lại có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\); \(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\); \(\frac{1}{4^2}>\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\); ...;\(\frac{1}{100^2}< \frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)

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

=> C > 99-1

=> C > 98

Ta có :C=

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

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

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

Mà \(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)l<\(\frac{100}{101}\)(tự tính)

Suy ra C> 98(đpcm)

C = 3/4 + 8/9 + 15/16 + ... + 9999/10000

C = 1- 1/4 + 1- 1/9 + 1- 1/16 + ... + 1- 1/10000

C = ( 1+1+1+...+1) - (1/2.2 + 1/3.3 + 1/4.4 + ...+ 1/100.100) >

(1+1+1+...+1) - ( 1/1.2+1/2.3+1/3.4+...+1/99.100) = 99 - ( 1/1-1/2+1/2-1/3+1/3+1/4+...+1/9999-1/10000

= 99 - ( 1-1/10000)= 99 - 1 + 1/10000= 98+1/10000 > 98

Vậy C > 98

Nhanh lên cac bạn ơi

mình ra là 0

A=3/4+8/9+15/16+...+9999/1000.

= 1 - 1/4 + 1  - 1/9 + 1 - 1/6 ... + 1 - 1/1000

= ( 1 + 1 + 1 + ... + 1 ) + ( - 1/4 - 1/6 - 1/9 - 1/1000 )

= 99 + (- 1/4 - 1/9 - 1/6 - ... - 1/1000 )

Vì 99 + ( - 1/4 - 1/9 = 1/6 - ... - 1/1000 )

=> A > 98

Vậy A > 98