Question

Chứng minh rằng\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{80^2}<1\)

Answer 1

Đặt A=\(\frac{1}{2^2}+\frac{1}{3^2}+..........+\frac{1}{80^2}\)

Ta có:\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};................;\frac{1}{80^2}<\frac{1}{79.80}\)

=>A\(<\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{79.80}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..............+\frac{1}{79}-\frac{1}{80}\)

=\(1-\frac{1}{80}<1\)

Vậy A<1(đpcm)

Answer 2

tacó 

1/2^2<1/1.2

1/3^2<1/2.3

1/4^2<1/3.4

.......………

1/80^2<1/79.80

->1/3^2+1/4^2+1/5^2+…+1/80^2< 1/1.2+ 1/2.3+1/3.4+1/4.5+..+1/79.80

->…................................................<1-1/2+1/2-1/3+1/3-1/4+…+1/79-1/80

->..................................................<1-1/80<1(₫pcm