Question

Answer 1

5a.

$M=1+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{97.99}+\frac{3}{99.101}$

$=1+\frac{3}{2}(\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{97.99}+\frac{2}{99.101})$

$=1+\frac{3}{2}(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101})$

$=1+\frac{3}{2}(\frac{1}{3}-\frac{1}{101})$

$=\frac{150}{101}$

Answer 2

Bài 5b.

$S=\frac{1}{4}(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2})$

$<\frac{1}{4}(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{(n-1)n})$

$=\frac{1}{4}(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n-1}-\frac{1}{n})$

$=\frac{1}{4}(1-\frac{1}{n})< \frac{1}{4}.1=\frac{1}{4}$
Ta có đpcm.