Question

Tính tổng :\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+........+\frac{19}{9^2.10^2}\)

 

Answer 1

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

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

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

= \(1-\frac{1}{100}=\frac{99}{100}\)

Answer 2

2/1×3  + 2/3×5  +2/5×7 +... +2/99×101

/ Là phần

Answer 3

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

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

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

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

Nhớ tk cho mk nha