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Ta có : \(\frac{1}{4.5}< \frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5.6}< \frac{1}{5^2}< \frac{1}{4.5}\)
.......
\(\frac{1}{99.100}< \frac{1}{99^2}< \frac{1}{98.99}\)
\(\frac{1}{101.100}< \frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}+\frac{1}{101.100}< A< \frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(\frac{1}{4}-\frac{1}{101}< A< \frac{1}{3}-\frac{1}{100}\Rightarrow\frac{97}{404}< A< \frac{97}{300}\)
=> A không phải là số tự nhiên ( đpcm )
\(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3C=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow C+3C=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4C< 1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}=D\)
Xét \(D=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(\frac{D}{3}=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(\Rightarrow D+\frac{D}{3}=1-\frac{1}{3^{100}}< 1\Rightarrow\frac{4D}{3}< 1\Rightarrow D< \frac{3}{4}\)
\(\Rightarrow4C< D< \frac{3}{4}\Rightarrow C< \frac{3}{16}\)
Ta có \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+....+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{1}{2}-\frac{1}{100}=\frac{49}{100}< \frac{3}{4}\left(đpcm\right)\)
\(C=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\)
\(\Rightarrow3C=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{99}{3^{98}}+\frac{100}{3^{99}}\)
Trừ dưới cho trên:
\(2C=1+\frac{2}{3}-\frac{1}{3}+\frac{3}{3^2}-\frac{2}{3^2}+\frac{4}{3^3}-\frac{3}{3^3}+...+\frac{100}{3^{99}}-\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(2C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
Đặt \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}=B\Rightarrow2C=B-\frac{100}{3^{100}}\)
\(B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3B=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3B-3+\frac{1}{3^{99}}=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}=B\)
\(\Rightarrow2B=3-\frac{1}{3^{99}}\Rightarrow B=\frac{3}{2}-\frac{1}{2.3^{99}}< \frac{3}{2}\)
\(\Rightarrow2C=B-\frac{100}{3^{100}}< B< \frac{3}{2}\Rightarrow C< \frac{3}{4}\)
b) A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
3A=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
3A-A=\(1-\frac{1}{3^{99}}\)
2A=\(1-\frac{1}{3^{99}}\)
vì 2A<1
=> A<\(\frac{1}{2}\)
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{2}{2}+\frac{3}{2^2}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)
\(2A-A=1+\frac{2}{2}-\frac{1}{2}+\frac{3}{2^2}-\frac{2}{2^2}+...+\frac{100}{2^{99}}-\frac{99}{2^{99}}-\frac{100}{2^{100}}\)
\(\Rightarrow A=2+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(\Rightarrow A=2.\frac{1}{2^{100}}\)
Vậy \(A< 2\) do \(A=2\) nhân với một phân số nhỏ hơn \(1\)
