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Sửa đề \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
Ta có: \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
\(=\left(2019+1\right)+\left(\frac{2018}{2}+1\right)+...+\left(\frac{1}{2019}+1\right)-2019\)
\(=2020+\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}-2020\)
\(=\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}\)
\(=2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)Thay vào biểu thức A ta được:
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{1}{2019}+\frac{2}{2018}+\frac{3}{2017}+...+\frac{2018}{2}+\frac{2019}{1}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{1}{2019}+1+\frac{2}{2018}+1+\frac{3}{2017}+1+...+\frac{2018}{2}+1+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{2020}{2019}+\frac{2020}{2018}+\frac{2020}{2017}+...+\frac{2020}{2}+\frac{2020}{2020}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{2020\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}\right)}\)
\(\frac{A}{B}=\frac{1}{2020}\)
Lời giải:
$D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+......+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}$
$4D=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}$
Trừ theo vế:
\(3D=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow 12D=4+1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2017}}-\frac{2019}{4^{2018}}\)
Trừ theo vế:
$9D=4-\frac{2019}{4^{2018}}+\frac{2019}{4^{2019}}-\frac{1}{4^{2018}}$
$=4-\frac{6061}{4^{2019}}< 4$
$\Rightarrow D< \frac{4}{9}<\frac{4}{8}$ hay $D< \frac{1}{2}$ (đpcm)
ta có B= 1/2018+2/2017+3/2016+...+2017/2+2018/1
=> B=1+1+1+..+1( 2018 số hạng 1)+ 1/2018+..+2017/2
=> B= (1+1/2018)+(1+2/2017)+(1+3/2016)+...+(1+2017/2)+ 2019/2019
=> B= 2019 *(1/2+1/3+...+1/2019)
=> A/B= (1/2+1/3+...+1/2019)/2019*(1/2+1/3+..+1/2019)
=> A/B= 1/2019
Ta có : \(\frac{x-1}{2017}+\frac{x-2}{2018}-\frac{x-3}{2019}=\frac{x-4}{2020}\)
\(\Rightarrow\frac{x-1}{2017}+\frac{x-2}{2018}=\frac{x-4}{2020}+\frac{x-3}{2019}\)
\(\Rightarrow1+\frac{x-1}{2017}+1+\frac{x-2}{2018}=1+\frac{x-4}{2020}+1+\frac{x-3}{2019}\)
\(\Rightarrow\frac{2016+x}{2017}+\frac{2016+x}{2018}=\frac{2016+x}{2020}+\frac{2016+x}{2019}\)
\(\Rightarrow\frac{2016+x}{2017}+\frac{2016+x}{2018}-\frac{2016+x}{2019}-\frac{2016+x}{2020}=0\)
\(\Rightarrow\left(2016+x\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
\(\text{Mà :
}\)\(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\)
\(\text{Nên : }\) \(2016+x=0\)
\(\Rightarrow x=-2016\)
\(D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}\)
\(\Rightarrow4D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)
\(\Rightarrow4D-D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)
\(-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-\frac{4}{4^4}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow3D=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}\right)-\frac{2019}{4^{2019}}\)
Đặt \(M=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+...+\frac{1}{4^{2018}}\)
\(\Rightarrow4M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(\Rightarrow4M-M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(-\frac{1}{4}-\frac{1}{4^2}-\frac{1}{4^3}-\frac{1}{4^4}-...-\frac{1}{4^{2018}}\)
\(\Rightarrow3M=1-\frac{1}{4^{2018}}\)
\(\Rightarrow M=\frac{1}{3}-\frac{1}{3.4^{2018}}\)
\(\Rightarrow3D=1+\frac{1}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow3D=\frac{4}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}< \frac{4}{3}\)
\(\Rightarrow D< \frac{4}{9}=\frac{40}{90}< \frac{45}{90}=\frac{1}{2}\left(đpcm\right)\)
\(MS=\frac{2019}{1}+\frac{2018}{2}+\frac{2017}{3}+...+\frac{1}{2019}\)
\(=1+1+\frac{2018}{2}+1+\frac{2017}{3}+...+1+\frac{1}{2019}\)
\(=\frac{2020}{2020}+\frac{2020}{2}+\frac{2020}{3}+...+\frac{2020}{2019}\)
\(=2020\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)
Suy ra \(D=\frac{1}{2020}\).