Question

chứng minh 3/1^2.2^2+5/2^2.3^2+...+19/9^2.10^2<1

Answer 1

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

=(1/1-1/4)+(1/4-1/9)+........+(1/81-1/100)

=1-1/100

=99/100<1(đpcm)

Answer 2

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

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

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

\(=1-\frac{1}{100}<1\)

Vậy _____________________