Violympic toán 6
Các câu hỏi tương tự
Cho A = \(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\). Chứng minh rằng A < \(\dfrac{7}{4}\)
Chứng minh rằng : A = \(\dfrac{1}{2}-\dfrac{2}{2^2}+\dfrac{3}{2^3}-\dfrac{4}{2^4}+....+\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}< \dfrac{2}{9}\)
Chứng minh rằng:
\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
Chứng minh rằng:
\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
chứng minh rằng:
E=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{3}{4}\)
Cho \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{2003^2}\)
Chứng minh rằng \(\dfrac{1}{3}< A< 1\)
Cho \(A=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2019}\)
Chứng minh A ko phải là số tự nhiên
20 tháng 3 2021 lúc 16:09
Giúp mk vớiCâu 1:Cho A dfrac{1}{dfrac{99}{dfrac{1}{2}+}}+dfrac{2}{dfrac{98}{dfrac{1}{3}+}}+dfrac{3}{dfrac{97}{dfrac{1}{4}+....}}+...+dfrac{99}{dfrac{1}{dfrac{1}{100}}}. B dfrac{92}{dfrac{1}{45}+}-dfrac{1}{dfrac{9}{dfrac{1}{50}+}}-dfrac{2}{dfrac{10}{dfrac{1}{55}+}}-dfrac{3}{dfrac{11}{dfrac{1}{60}+....}}-...dfrac{92}{dfrac{100}{dfrac{1}{500}}}. Tính dfrac{A}{B}
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Giúp mk với
Câu 1:
Cho A = \(\dfrac{1}{\dfrac{99}{\dfrac{1}{2}+}}+\dfrac{2}{\dfrac{98}{\dfrac{1}{3}+}}+\dfrac{3}{\dfrac{97}{\dfrac{1}{4}+....}}+...+\dfrac{99}{\dfrac{1}{\dfrac{1}{100}}}\).
B =\(\dfrac{92}{\dfrac{1}{45}+}-\dfrac{1}{\dfrac{9}{\dfrac{1}{50}+}}-\dfrac{2}{\dfrac{10}{\dfrac{1}{55}+}}-\dfrac{3}{\dfrac{11}{\dfrac{1}{60}+....}}-...\dfrac{92}{\dfrac{100}{\dfrac{1}{500}}}\). Tính \(\dfrac{A}{B}\)
Cho S=\(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2023}{4^{2023}}\). Chứng minh S < \(\dfrac{1}{2}\)