\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=\frac{9999}{10000}\)

 \(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=\frac{9999}{10000}\)

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tick nhaTran Thai Han Thuyen

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tick nha

\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}\)

= \(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)

= \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}\)

= \(1-\frac{1}{100}=\frac{99}{100}\)

2/1×3  + 2/3×5  +2/5×7 +... +2/99×101

/ Là phần

\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2+10^2}\)

\(=>\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)

\(=>1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)

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

Nhớ tk cho mk nha

3/1².2² + 5/2².3² + 7/3².4² + ... + 19/9².10²

= 3/1.4 + 5/4.9 + 7/9.16 + ... + 19/81.100

= 1 - 1/4 + 1/4 - 1/9 + 1/9 - 1/16 + ... + 1/81 - 1/100

= 1 - 1/100

= 99/100

\(A=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+..+\frac{17}{64.81}+\frac{19}{81.100}\)

\(A=\frac{4-1}{1.4}+\frac{9-4}{4.9}+\frac{16-9}{9.16}+...+\frac{81-64}{64.81}+\frac{100-81}{81.100}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{61}-\frac{1}{81}+\frac{1}{81}-\frac{1}{100}\)

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

\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+.......+\frac{10^2-9^2}{9^2.10^2}\)

    \(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+.....+\frac{1}{9^2}-\frac{1}{10^2}\)

   \(=\frac{1}{1^2}-\frac{1}{10^2}=1-\frac{1}{100}=\frac{99}{100}\)

\(H=\frac{8}{1^2\cdot3^2}+\frac{16}{3^2\cdot5^2}+...+\frac{48}{11^2\cdot13^2}\)

\(H=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{5^2}+...+\frac{1}{11^2}-\frac{1}{13^2}\)

\(H=1-\frac{1}{13^2}\)

\(H=\frac{168}{169}\)

Phương thiếu bước nhé 

\(H=\frac{8}{1^2.3^2}+\frac{16}{3^2.5^2}+\frac{24}{5^2.7^2}+...+\frac{48}{11^2.13^2}\)

\(H=\frac{3^2-1^2}{1^2.3^2}+\frac{5^2-3^2}{3^2.5^2}+\frac{7^2-5^2}{5^2.7^2}+...+\frac{13^2-11^2}{11^2.13^2}\)

\(H=\frac{3^2}{1^2.3^2}-\frac{1^2}{1^2.3^2}+\frac{5^2}{3^2.5^2}-\frac{3^2}{3^2.5^2}+\frac{7^2}{5^2.7^2}-\frac{5^2}{5^2.7^2}+...+\frac{13^2}{11^2.13^2}-\frac{11^2}{11^2.13^2}\)

\(H=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{5^2}+\frac{1}{5^2}-\frac{1}{7^2}+...+\frac{1}{11^2}-\frac{1}{13^2}\)

\(H=1-\frac{1}{13^2}=1-\frac{1}{169}=\frac{168}{169}\)

Chúc bạn học tốt ~ 

< 1 nhé 

Ta có: \(\frac{3}{1^2.2^2}=\frac{3}{1.4}=1-\frac{1}{4}\); \(\frac{5}{2^2.3^2}=\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\); \(\frac{7}{3^2.4^2}=\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\); ...; \(\frac{39}{19^2.20^2}=\frac{39}{361.400}=\frac{1}{361}-\frac{1}{400}\)

Gọi tổng đó là A => A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{361}-\frac{1}{400}\)

=> \(A=1-\frac{1}{400}=\frac{399}{400}< \frac{400}{400}=1\)

=> A < 1

3/1^2.2^2+5/2^2.3^2+7/3^2.4^2+...+4019/2009^2.2010^2

=3/1.4+5/4.9+7/9.16+...+4019/4036081.4040100

= 1/1-1/4+1/4-1/9+1/9-1/16+...+1/4036081-1/4040100

= 1/1-1/4040100

= 1-1/4040100 < 1

Chúc bạn học tốt!

câu hỏi là tính tổng à

k bạn ơi nó là cm  cái tổng đó <1