mấy bạn ơi câu b) là chứng minh C<\(\dfrac{1}{2}\)nha

\(S=\dfrac{1}{2018}\left(1+\dfrac{1}{1}+1+\dfrac{1}{2}+1+\dfrac{1}{3}+...+1+\dfrac{1}{2018}\right)\)

\(S=\dfrac{1}{2018}\left(2018+\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)

\(S=1+\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)

Do \(\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2018}\right)>0\Rightarrow S>1\) (1)

Lại có:

\(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}< \dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}+...+\dfrac{1}{1}=2018\)

\(\Rightarrow1+\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)< 1+\dfrac{1}{2018}.2018=2\)

\(\Rightarrow S< 2\) (2)

Từ (1), (2) \(\Rightarrow1< S< 2\)

\(\Rightarrow S\) nằm giữa 2 số tự nhiên liên tiếp nên S không phải là số tự nhiên

S=1/2018

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2018}}+\dfrac{1}{3^{2019}}\)

\(3A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2017}}+\dfrac{1}{3^{2018}}\)

\(3A-A=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{2018}}\right)-\left(\dfrac{1}{3}+...+\dfrac{1}{3^{2019}}\right)\)

\(2A=1-\dfrac{1}{3^{2019}}\)

\(A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{2019}}< \dfrac{1}{2}\) (DPCM)

\(3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2017}}+\dfrac{1}{3^{2018}}\)

\(\Rightarrow3A-A=1-\dfrac{1}{3^{2019}}\)

\(\Rightarrow2A=1-\dfrac{1}{3^{2019}}\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{2019}}\)

\(\Rightarrow A< \dfrac{1}{2}\)

Theo bài ra, ta có: \(B=\dfrac{2018}{1}+\dfrac{2017}{2}+\dfrac{2016}{3}+...+\dfrac{1}{2018}\)

\(B=\left(\dfrac{2018}{1}+1\right)+\left(\dfrac{2017}{2}+1\right)+\left(\dfrac{2016}{3}+1\right)+...+\left(\dfrac{1}{2018}+1\right)-2018\)

\(B=2019+\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}-2018\)

\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\left(2019-2018\right)\)

\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+1\)

\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\dfrac{2019}{2019}\)

\(B=2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)\)

Khi đó:\(\dfrac{B}{A}=\dfrac{2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}}\)

\(\Rightarrow\dfrac{B}{A}=2019\), là 1 số nguyên.

Vậy \(\dfrac{B}{A}\) là số nguyên.

\(A=1.2.3...2018\left[\left(1+\dfrac{1}{2018}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2017}\right)+...+\left(\dfrac{1}{1009}+\dfrac{1}{1010}\right)\right]\)

\(A=1.2.3...2018.2019\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\)

\(\dfrac{A}{2019}=1.2.3...2018\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\).

Rõ ràng tích 1 . 2 ... 2018 chia hết cho các tích 1 . 2018; 2 . 2017; ...; 1009 . 1010; do đó \(\dfrac{A}{2019}\) là số tự nhiên.

Vậy A chia hết cho 2019.

a) \(2\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+...+\dfrac{16}{n\left(n+16\right)}\right)=\dfrac{16}{25}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+16}=\dfrac{8}{25}\)

\(\dfrac{1}{3}-\dfrac{1}{n+16}=\dfrac{8}{25}\)

\(\dfrac{n+13}{3\left(n+16\right)}=\dfrac{8}{25}\)

\(24n+384=25n+325\)

\(25n-24n=384-325\)

\(n=59\)

b) Sai đề nha

\(\left\{{}\begin{matrix}\dfrac{2018}{2019}< 1\\\dfrac{2019}{2020}< 1\\\dfrac{2020}{2021}< 1\\\dfrac{2021}{2022}< 1\end{matrix}\right.\)

\(\Rightarrow\dfrac{2018}{2019}+\dfrac{2019}{2020}+\dfrac{2020}{2021}+\dfrac{2021}{2022}< 4\)

chị ơi hình như chị nhầm rồi P/s cuối phải là 1/n.(n+6)thì phải

Ta có :

B = \(\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)

B = \(\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)

B = \(\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+1\)

B = \(2021\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+...+\dfrac{1}{2}\right)\)  (1)

Mà A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\)   (2)

Từ (1) và (2) \(\Rightarrow\) \(\dfrac{A}{B}=\dfrac{1}{2021}\)

Ta có: \(B=\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)

\(=\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)

\(=\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+\dfrac{2021}{2021}\)

Suy ra: \(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}}{2021\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=\dfrac{1}{2021}\)

Giải:

Ta có:

\(B=\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\) 

\(B=1+\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)\) 

\(B=\dfrac{2021}{2021}+\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}\) 

\(B=2021.\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+...+\dfrac{1}{2}\right)\) 

\(\Rightarrow\dfrac{A}{B}=\dfrac{\left[2021.\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+...+\dfrac{1}{2}\right)\right]}{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=2021\) 

Vậy \(\dfrac{A}{B}=2021\)

\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)

\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)

=>B<1

=>A>B