Chứng tỏ A < \(\frac{9}{2}\)

giúp mình đi 

Đặt A=\(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+\frac{19}{3^4}+...+\frac{2015}{3^{503}}+\frac{2019}{3^{504}}\)

3A=\(7+\frac{11}{3}+\frac{15}{3^2}+\frac{19}{3^3}+...+\frac{2015}{3^{502}}+\frac{2019}{5^{503}}\)

=> 3A-A=(\(7+\frac{11}{3}+\frac{15}{3^2}+\frac{19}{3^3}+...+\frac{2015}{3^{502}}+\frac{2019}{5^{503}}\))-(\(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+\frac{19}{3^4}+...+\frac{2015}{3^{503}}+\frac{2019}{3^{504}}\))

2A=\(7+\left(\frac{11}{3}-\frac{7}{3}\right)+\left(\frac{15}{3^2}-\frac{11}{3^2}\right)+\left(\frac{19}{3^3}-\frac{15}{3^3}\right)+...+\left(\frac{2019}{3^{503}}-\frac{2015}{3^{503}}\right)-\frac{2019}{3^{504}}\)

2A=\(7+\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+...+\frac{4}{3^{503}}-\frac{2019}{3^{504}}\)

=> A=\(\frac{7}{2}+2\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{503}}\right)-\frac{2019}{2.3^{504}}\)

Em làm tiếp Xét 

B=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{503}}\)

3B=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{502}}\)

=> 3B-B=\(1-\frac{1}{3^{503}}\)

=> B=\(\frac{1}{2}-\frac{1}{2.3^{503}}\)

=> A=\(\frac{7}{2}+2\left(\frac{1}{2}-\frac{1}{2.3^{503}}\right)-\frac{2019}{2.3^{504}}=\frac{9}{2}-\frac{1}{3^{503}}-\frac{2019}{2.3^{504}}< \frac{9}{2}\)

<3 <3 <3

\(A=2\cdot\left(\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2017^2}\right)< 2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}\right)\)

Đặt \(M=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}=\left(1+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)

\(\Rightarrow M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)

\(\Rightarrow M=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}< \frac{1}{1009}+\frac{1}{1009}+...+\frac{1}{1009}\)(1008 số hạng )

hay\(M< \frac{1008}{1009}\Rightarrow A< 2\cdot\frac{1008}{1009}=\frac{504}{1009}\left(ĐPCM\right)\)

ta có:   L = \(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+...+\frac{403}{3^{100}}\)

<=> \(3L=7+\frac{11}{3}+\frac{15}{3^2} +..+\frac{403}{3^{99}}\)

=> \(3L-L=\left(7+\frac{11}{3}+\frac{15}{3^2}+...+\frac{403}{3^{99}}\right)-\left(\frac{7}{3}+\frac{11}{3^2}+...+\frac{403}{3^{100}}\right)\)

<=> \(2L=7+\left(\frac{11}{3}-\frac{7}{3}\right)+\left(\frac{15}{3^2}-\frac{11}{3^2}\right)+...+\left(\frac{403}{3 ^{99}}-\frac{399}{3^{99}}\right)-\frac{403}{3^{100}}\)

<=> \(2L=7+4\cdot\frac{1}{3}+4\cdot\frac{1}{3^2}+..+4\cdot\frac{1}{3^{99}}-\frac{403}{3^{100}}\)

<=> \(2L=7+4\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{403}{3^{100}}\)

<=>\(2L=7+4\left[\frac{1}{2}\cdot\left(1-\frac{1}{3^{99}}\right)\right]-\frac{403}{3^{100}}\)

<=> \(2L=7+2-\frac{2}{3^{99}}-\frac{403}{3^{100}}\)

<=> \(L=3,5+1-\frac{1}{3^{99}}-\frac{403}{2\cdot3^{100}}\)

<=> \(L=4,5-\frac{1}{3^{99}}-\frac{403}{2\cdot3^{100}}

a) \(A=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)

\(\Rightarrow3A=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{100}}\)

\(\Rightarrow3A-A=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{99}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+...+\frac{301}{3^{100}}\right)\)

\(\Rightarrow2A=4+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{301}{3^{100}}\)

Đặt \(F=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)

\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)

\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)

\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)

\(\Rightarrow F< \frac{3}{2}\)

\(\Rightarrow2A< 4+\frac{3}{2}\)

\(\Rightarrow2A< \frac{11}{2}\)

\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)

2. \(B=\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)

\(\Rightarrow3B=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\)

\(\Rightarrow3B-B=\left(11+...+\frac{605}{3^{99}}\right)-\left(\frac{11}{3}+...+\frac{605}{3^{100}}\right)\)

\(\Rightarrow2B=11+2+\frac{2}{3}+...+\frac{2}{3^{98}}-\frac{605}{3^{100}}\)

Đặt \(D=2+\frac{2}{3}+...+\frac{2}{3^{98}}\)

\(\Rightarrow3D=6+2+...+\frac{2}{3^{97}}\)

\(\Rightarrow2D=6-\frac{2}{3^{98}}< 6\)( làm tắt )

\(\Rightarrow2D< 6\)

\(\Rightarrow D< 3\)

\(\Rightarrow2B< 11+3\)

\(\Rightarrow2B< 14\)

\(\Rightarrow B< 7\left(đpcm\right)\)

Phần cuối cũng tương tự 2 phần mình vừa làm nhé

Bạn tự làm nốt nhé đánh mệt lắm

Ta có bài toán tổng quát sau:Chứng minh rằng tổng \(A=\frac{n+1}{n^2+1}+\frac{n+1}{n^2+2}+....+\frac{n+1}{n^2+n}\)(n số hạng và n>1) không phải là số nguyên dương ta có:

\(1=\frac{n+1}{n^2+1}+\frac{n+1}{n^2+2}+...+\frac{n+1}{n^2+3}< \frac{n+1}{n^2+1}+\frac{n+1}{n^2+2}+....+\frac{n+1}{n^2+n}< \frac{n+1}{n^2}+\frac{n+1}{n^2}\)\(+....+\frac{n+1}{n^2}=2\)

Do đó A không phải là số nguyên dương với n=2019 thì ta có bài toán đã cho

làm ơn

bài này lm kiểu j vậy bạn