Question

Tính

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

Help me...

Answer 1

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

Answer 2

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

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

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

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