Giải:

\(S=\dfrac{1}{2}+\dfrac{2}{2^2}+...+\dfrac{n}{2^n}+...+\dfrac{2017}{2^{2017}}\) 

Với \(n>2\) thì \(\dfrac{n}{2^n}=\dfrac{n+1}{2^{n-1}}-\dfrac{n+2}{2^n}\) 

Ta có:

\(\dfrac{n+1}{2^{n-1}}=\dfrac{n+1}{2^n:2}=\dfrac{2.\left(n+1\right)}{2^n}\) 

\(\Rightarrow\dfrac{n+1}{2^{n-1}}-\dfrac{n+2}{2^n}\) 

\(=\dfrac{2.\left(n+1\right)}{2^n}-\dfrac{n+2}{2^n}\) 

\(=\dfrac{2.\left(n+1\right)-n-2}{2^n}\) 

\(=\dfrac{n}{2^n}\) 

  \(\Leftrightarrow S=\dfrac{1}{2}+\left(\dfrac{2+1}{2^{2-1}}-\dfrac{2+2}{2^2}\right)+...+\left(\dfrac{2016+1}{2^{2015}}-\dfrac{2018}{2^{2016}}\right)+\left(\dfrac{2017+1}{2^{2016}}-\dfrac{2019}{2^{2017}}\right)\)

\(S=\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{2019}{2017}\) 

\(S=2-\dfrac{2019}{2017}\)  

\(\Leftrightarrow S=2-\dfrac{2019}{2017}< 2\) 

Hay \(S< 2\)

Cảm ơn bạn ^-^

Ta có: \(A=\frac{2^{2017}+2}{2^{2017}+3}=1-\frac{1}{2^{2017}+3}\)

        \(B=\frac{2^{2017}+1}{2^{2017}+2}=1-\frac{1}{2^{2017}+2}\)

Vì \(\frac{1}{2^{2017}+3}< \frac{1}{2^{2017}+2}\) nên \(1-\frac{1}{2^{2017}+3}>1-\frac{1}{2^{2017}+2}\)

hay A > B

A>B bạn nhé

Ta có: \(T=\frac{2}{2^1}+\frac{3}{2^2}+\cdots+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\)

=>2T=\(2+\frac32+\cdots+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\)

=>2T-T=\(2+\frac32+\cdots+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}-\frac{2}{2^1}-\frac{3}{2^2}-\cdots-\frac{2016}{2^{2015}}-\frac{2017}{2^{2016}}\)

=>T=\(2+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{2015}}-\frac{2017}{2^{2016}}\)

Đặt \(A=\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{2015}}\)

=>2A=\(1+\frac12+...+\frac{1}{2^{2014}}\)

=>2A-A=\(1+\frac12+\cdots+\frac{1}{2^{2014}}-\frac12-\frac{1}{2^2}-\cdots-\frac{1}{2^{2015}}\)

=>\(A=1-\frac{1}{2^{2015}}=\frac{2^{2015}-1}{2^{2015}}\)

Ta có: \(T=2+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{2015}}-\frac{2017}{2^{2016}}\)

\(=2+\frac{2^{2015}-1}{2^{2015}}-\frac{2017}{2^{2016}}=2+\frac{2^{2016}-2-2017}{2^{2016}}=2+1-\frac{2019}{2^{2016}}<3\)

=>T<3

hình như là ko

Đề có đúng không thế bạn 

1 lần đúng 2

S<2²⁰¹⁷

nhé bạn