Lời giải:

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

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

$=99-(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+....+\frac{1}{10000})< 99$

Tham khảo :
 

3​.98​.1615​.....100009999​

=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{99.101}{100.100}=2.21.3​.3.32.4​.4.43.5​.....100.10099.101​

=\dfrac{\left(1.2.3.....99\right)}{\left(2.3.4.....100\right)}.\dfrac{\left(3.4.5.....101\right)}{\left(2.3.4.....100\right)}=(2.3.4.....100)(1.2.3.....99)​.(2.3.4.....100)(3.4.5.....101)​

=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}=1001​.2101​=200101​
 

cộng mà bạn

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)

Ta có: \(A=\left\{\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{9999}{10000}\right\}\Rightarrow99\)số

\(A=\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{9}\right)+...+\left(1-\dfrac{1}{100000}\right)\)

\(A=\left\{1+1+1+...+1\right\}\Rightarrow99\)số \(-\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{100000}=99-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{10000}\right)\)

Ta có: \(4=2^2>1.2\Rightarrow\dfrac{1}{4}< \dfrac{1}{1.2}\Leftrightarrow\dfrac{1}{4}< \dfrac{1}{1}-\dfrac{1}{2}\)

Tương tự: \(\dfrac{1}{9}< \dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{16}< \dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{10000}< \dfrac{1}{99}-\dfrac{1}{100}\)

Cộng theo vế ta được: \(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{10000}< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)

\(\Rightarrow A=99-\left(\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{16}+...+\dfrac{1}{10000}\right)>99-1=98\)

Vậy \(A>98\)

nhận xét: với n là số tự nhiên, ta có (n-1)(n+1)=n(n+1)-(n+1)=n²+n-n-1=n²-1

do đó: 1.3=2²-1

           2.4=3²-1

            ........

           99.101=100²-1

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

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

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

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

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

Ta có:

 \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}

chẳng hiểu gì cả

đúng ko vậy

khó ợt ( dễ ợt )

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 

chịu mẹ kiếp toán 7 cho vào đề kiểm tra toán 6 ai mà lm dc

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

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

=99+(-1/4-1/9-1/16-...-1/10000)

Vì 99+(-1/4-1/9-1/16-...-1/10000)>98

=>C>98

Vây C>98

Bài này lp 6 hc rùi mà