1/51+1/52+....+1/99<1/2

a < b k mình nha xong mình k lại cho

a) 

Ta có a > b vì b > 3 còn a < 3

 b)

a. Ta có : 1/51 + 1/52 + 1/53 +...+ 1/60 < 1/51 x 10 < 1/50 x 10 = 1/5

=> 1/51 + 1/52 +1/53 +...+1/60 < 1/5

b. Ta có : 1/51 + 1/52 + 1/53 +...+ 1/60 > 1/60 x 10 = 1/6

=> 1/51 + 1/52 +1/53 +...+ 1/60 > 1/6

a﴿
Ta có a > b vì b > 3 còn a < 3
b﴿
a. Ta có : 1/51 + 1/52 + 1/53 +...+ 1/60 < 1/51 x 10 < 1/50 x 10 = 1/5
=> 1/51 + 1/52 +1/53 +...+1/60 < 1/5
b. Ta có : 1/51 + 1/52 + 1/53 +...+ 1/60 > 1/60 x 10 = 1/6
=> 1/51 + 1/52 +1/53 +...+ 1/60 > 1/6

nhanh đi

Ta có : 

\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\) 

\(T=1+\frac{3}{1.2^2}+\frac{4}{2.2^2}+\frac{5}{2^2.2^2}+...+\frac{2016}{2^{2013}.2^2}+\frac{2017}{2^{1014}.2^2}\)

\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{4}+\frac{6}{8}+...+\frac{2016}{x}+\frac{2017}{x}\right)\)

\(=1+\frac{1}{2^2}.\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2013}}+\frac{2017}{2^{2014}}\right)\)

Đến chỗ này chịu!

Ta có

\(T=1+\frac{3}{1\cdot2^2}+\frac{4}{2\cdot2^2}+...+\frac{2017}{2^2\cdot2^{2014}}\) 

\(T=1+\frac{1}{2^2}\cdot\left(3+2+\frac{5}{2^2}+\frac{6}{2^3}+...+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\right)\)