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Chứng minh rằng:a. frac{1}{3^2}+frac{2}{3^3}+frac{3}{3^4}+frac{4}{3^5}+...+frac{99}{3^{100}}+frac{100}{3^{101}} frac{1}{4}b.frac{1}{2}-frac{1}{4}+frac{1}{8}-frac{1}{16}+frac{1}{32}-frac{1}{64} frac{1}{3}c.frac{1}{3}-frac{2}{3^2}+frac{3}{3^3}-frac{4}{3^4}+...+frac{99}{3^{99}}-frac{100}{3^{100}} frac{1}{16}d. frac{1}{5^2}-frac{2}{5^3}+frac{3}{5^4}-frac{4}{5^5}+...+frac{99}{5^{100}}-frac{100}{5^{101}} frac{1}{36}
Đọc tiếp
Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
Cho biểu thức \(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
Chứng minh rằng \(C< \frac{3}{16}\)
Cho biểu thức \(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
Chứng minh rằng \(C< \frac{3}{16}\)
2 tháng 4 2020 lúc 15:31
\(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
Hãy chứng minh: \(C< \frac{3}{16}\)
Cứu mị!
Chứng tỏ rằng \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Cho C = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)chứng minh rằng C < \(\frac{1}{2}\)CMR : \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)CMR : \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
Xem chi tiết
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
CMR: 100-(\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\))=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
CMR:
a,\(100\left(1+\frac{1}{2}+\frac{1}{3}+..........+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+........+\frac{99}{100}\)