A = ( 1 + 2^1 ) + ( 2^2 + 2^3 ) + ... + ( 2^10 + 2^11 )

A = 3 . 1 + 3 . 4 + ... + 3 . 1024

A = 3 ( 1 + 4 + ... + 1024 )

=> A chia hết cho 3 

\(A=\left(1+2\right)+\left(2^2+2^3\right)+...........+\left(2^{10}+2^{11}\right)\)

\(=3+2^2.3+.............+2^{10}.3\)

\(=\left(1+2^2+........+2^{10}\right).3\) chia hết cho 3

Vậy A chia hết cho 3

SAI ĐỀ RỒI

sorry mina , phải là 1/10²+11²

sai chỗ nào vậy P.H.Đăng

đặt tổng trên là A

ta co:

1/10^2<1/9.10

1/11^2<1/10.11

........

1/2014^2

=>A<1/9.10+1/10.11+.......+1/2013.2014=1/9-1/2014<1/9<9(đpcm)

 P = 1/5^2 + 2/5^3 + 3/5^4 + ... + 10/5^11 + 11/5^12 . 

5P = \(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{10}{5^{10}}+\frac{11}{5^{11}}\)

5P - P = ( \(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{10}{5^{10}}+\frac{11}{5^{11}}\)) - ( 1/5^2 + 2/5^3 + 3/5^4 + ... + 10/5^11 + 11/5^12 .  )

4P = \(\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)-\frac{11}{5^{12}}\)

4P = \(\frac{1-\frac{1}{5^{11}}}{4}-\frac{11}{5^{12}}< \frac{1}{4}\)

\(P< \frac{1}{16}\)

\(A=\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{199}+\dfrac{1}{200}\\ >\dfrac{1}{10}+\left(\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}\right)\left(90so\right)+\left(\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\right)\left(100so\right)\\ A>\dfrac{1}{10}+\dfrac{90}{100}+\dfrac{100}{200}=1+\dfrac{1}{2}=\dfrac{3}{2}\left(đpcm\right).\)