A=1/2^{2 +} 1/3² + 1/4² + ... + 1/100²+1/1990²

Có 1/3²<1/2.3

...

...

1/1990²<1/1989.1990

⇒A=1/2^{2 +} 1/3² + 1/4² + ... + 1/100²<1/4+1/2x3+1/3x4+..+1/1989x1990

⇒A<1/4+1/2-1/3+1/3-1/4+..+1/989-1/1990=1/4+1/2-1/1990=3/4-1/1990<3/4

Chúc em học tốt

Nocopy

Xin câu trả lời hay nhất

$\dfrac{1}{2^2}<\dfrac{1}{2.3}$

$\dfrac{1}{3^2}<\dfrac{1}{3.4}$

.............

$\dfrac{1}{1990^2}<\dfrac{1}{1990.1991}$

$→A<\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{1990.1991}$

$→A<\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1990}-\dfrac{1}{1991}$

$→A<\dfrac{1}{2}-\dfrac{1}{1991}$

$→A<\dfrac{1989}{3982}≈0,5$

mà $\dfrac{1989}{3982}<\dfrac{3}{4}$

$→A<\dfrac{3}{4}$

$→DPCM$