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!

A có : 100 - 2 + 1 = 99 thừa số.

Tất cả thừa số của A đều âm.

=> A < 0 < \(\frac{1}{2}\)

\(M=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}\)

\(>1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)

\(=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

\(=1+\frac{1}{2}-\frac{1}{11}\)

\(>1+\frac{1}{2}-\frac{1}{6}=\frac{4}{3}\)

\(Q=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{40^2}\right)\)

\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)...\left(\frac{40^2-1}{40^2}\right)\)

\(=\frac{\left(2-1\right)\left(2+1\right)}{2^2}\frac{\left(3-1\right)\left(3+1\right)}{3^2}...\frac{\left(40-1\right)\left(40+1\right)}{40^2}\)

\(=\frac{1.3.2.4.3.5...39.41}{2^2.3^2.4^2...40^2}\)

\(=\frac{1.2.3...39}{2.3.4...40}.\frac{3.4.5...41}{2.3.4...40}=\frac{1}{40}.\frac{41}{2}=\frac{1}{2}.\frac{41}{40}\)

Mà \(41>40\Rightarrow\frac{41}{40}>1\Rightarrow\frac{1}{2}.\frac{41}{40}>\frac{1}{2}\Rightarrow A>\frac{1}{2}\)

Cảm ơn bạn nhé!!!