Chứng tỏ a/D = 1/101 + 1/102 + 1/103 + … +1/200 > 1/2 b/N = 1/2^2 + 1/3^2 + 1/4^2 + … + 1/100^2 < 1

Giải thích các bước giải:

a.Ta có :
$D=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+…+\dfrac{1}{200}$ 

$\to D>\dfrac{1}{200}+\dfrac{1}{200}+\dfrac{1}{200}+…+\dfrac{1}{200}$  có 100 số hạng

$\to D>\dfrac{100}{200}$

$\to D>\dfrac12$

b.Ta có :
$N=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+..+\dfrac{1}{100^2}$

$\to N=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+..+\dfrac{1}{100.100}$

$\to N<\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{99.100}$

$\to N<\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+..+\dfrac{100-99}{99.100}$

$\to N<\dfrac11-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14+..+\dfrac1{99}-\dfrac1{100}$

$\to N<1-\dfrac1{100}$

$\to N<1$