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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<\(\frac{1}{2}\)
Đặt \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(\Rightarrow A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(\Rightarrow A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{100}{100!}-\frac{1}{100!}\)
\(\Rightarrow A=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(\Rightarrow A=\frac{1}{1!}-\frac{1}{100!}\)
\(\Rightarrow A=1-\frac{1}{100!}\)
Mà \(1-\frac{1}{100!}< 1.\)
\(\Rightarrow A< 1\left(đpcm\right).\)
Chúc bạn học tốt!
Chứng minh rằng: P = \(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}< \frac{5}{16}\)
\(P=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)
\(5P=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{99}{5^{98}}\)
\(\Rightarrow4P=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}=A-\frac{99}{5^{99}}\)
\(A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}\)
\(5A=5+1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{97}}\)
\(\Rightarrow4A=5-\frac{1}{5^{98}}< 5\Rightarrow A< \frac{5}{4}\)
\(4P=A-\frac{99}{5^{99}}< A< \frac{5}{4}\Rightarrow P< \frac{5}{16}\)
Câu hỏi của Ngô Văn Nam - Toán lớp 6 - Học toán với OnlineMath
Ta có: \(y=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}\Leftrightarrow3y=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{98}}\)
\(\Leftrightarrow3y-y=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{99}\right)\)
\(\Leftrightarrow2y=1-\frac{1}{3^{99}}
\(Q=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(3Q=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3Q-Q=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(2Q=1-\frac{1}{3^{100}}< 1\)
\(\Rightarrow Q=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1}{2}\)