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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=\(\frac{7}{2}+2\left(\frac{1}{2}-\frac{1}{2.3^{503}}\right)-\frac{2019}{2.3^{504}}=\frac{9}{2}-\frac{1}{3^{503}}-\frac{2019}{2.3^{504}}< \frac{9}{2}\)
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 )
\(\Rightarrow2D< 6\)
\(\Rightarrow D< 3\)
\(\Rightarrow2B< 11+3\)
\(\Rightarrow2B< 14\)
\(\Rightarrow B< 7\left(đpcm\right)\)
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