Ta có :
\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{19}{\left(9.10\right)^2}\)
\(=\)\(\frac{3}{1.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\)\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\)\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=\)\(1-\frac{1}{100}\)
\(=\)\(\frac{100}{100}-\frac{1}{100}\)
\(=\)\(\frac{100-1}{100}\)
\(=\)\(\frac{99}{100}\)
Vậy ...
Đặt A=\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+..........+\frac{19}{\left(9.10\right)^2}\)
A=\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+.........+\frac{19}{9^2.10^2}\)
A=\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...........+\frac{19}{81.100}\)
A=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-...............+\frac{1}{81}-\frac{1}{100}\)
A=\(\frac{1}{1}-\frac{1}{100}\)
A=\(\frac{99}{100}\)
Vậy tổng của \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+..........+\frac{19}{\left(9.10\right)^2}\)là \(\frac{99}{100}\)
Chúc bn học tốt
