khó quá 

Ta có: \(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

\(\frac{1}{4^2}<\frac{1}{3\cdot4}=\frac13-\frac14\)

...

\(\frac{1}{20^2}<\frac{1}{19\cdot20}=\frac{1}{19}-\frac{1}{20}\)

Do đó: \(\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{20^2}<\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{19}-\frac{1}{20}<\frac12\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{20^2}<\frac14+\frac12\)

=>\(A<\frac34\)

a)S=1+2+2^2+2^3+...+2^9

2S=2+2^2+2^3+...+2^10

2S-S=(2+2^2+2^3+2^4+...+2^10)-(1+2+2^2+2^3+...+2^9)

S=2^10-1

S=1024-1

S=1023

Ta có:5.2^8=5.256=1280

Mà 1280>1023

=>S<5.2^8

b)Ta có:M=1+2+2^2+2^3+2^4

=>2M=2+2^2+2^3+2^4+2^5

=>2M-M=(2+2^2+2^3+2^4+2^5)-(1+2+2^2+2^3+2^4)

=>M=2^5-1

Mà N=2^5-1

=>M=N

Không biết có bị sai lỗi nào hay không,nhớ kiểm tra đó

A = 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/2020^2

1/2^2 = 1/2.2 < 1/1.2 = 1/1 - 1/2

1/3^2 = 1/3.3 < 1/2.3 = 1/2 - 1/3

1/4^2 = 1/4.4 < 1/3.4 = 1/3 - 1/4

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

1/2020^2 < 1/2019.2020 = 1/2019 - 1/2020

Cộng vế với vế ta có:

A = 1/2^2 + 1/3^2+..+1/2020^2 = 1/1 - 1/2020

A < 1

1 < 3/2

Vậy A < 1 < 3/2

A=\(\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.....\frac{2016^2-1}{2016^2}\)

A=\(\frac{\left(2+1\right)\left(2-1\right)}{2^2}.\frac{\left(3+1\right)\left(3-1\right)}{3^2}......\frac{\left(2016+1\right)\left(2016-1\right)}{2016^2}\)

A=\(\frac{3.4......2017}{2.3....2016}.\frac{1.2...2015}{2.3...2016}\)

A=\(\frac{2017}{2}.\frac{1}{2016}\)

A=\(\frac{2017}{2.2106}>\frac{1}{2}\)

Vậy A\(>\frac{1}{2}\)