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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 \(C< \frac{3}{16}\)
chứng tỏ rằng
C = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}< \frac{1}{3}\)
D = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
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: \(C< \frac{3}{16}\)
cho bt: \(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}.\)CMR: C<\(\frac{3}{16}\)
Chứng tỏ rằng:
a) \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{17}{8^2.9^2}+\frac{19}{9^2.10^2}< 1\) 1
b) \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}< \frac{3}{4}\)
Giúp mik với, sáng mai 8h00 mik cần gấp. Bạn nào nhanh với trình bày đầy đủ mik tick cho ~_~
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
Afrac{1}{3}-frac{2}{3^2}+frac{3}{3^3}-...+frac{99}{3^{99}}-frac{100}{3^{100}}Rightarrow3A1-frac{2}{3}+frac{3}{3^2}-...+frac{99}{3^{98}}-frac{100}{3^{99}}Rightarrow3A+Aleft(...right)+left(...right)Rightarrow4A1-frac{1}{3}+frac{1}{3^2}-...-frac{1}{3^{99}}-frac{100}{3^{100}}Rightarrow3.4A3-1+frac{1}{3}-...-frac{1}{3^{98}}-frac{100}{3^{99}}Rightarrow12A+4Aleft(...right)+left(...right)Rightarrow16A3-frac{101}{3^{99}}-frac{100}{3^{100}} 3Rightarrow A frac{3}{16}
Đọc tiếp
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow3A+A=\left(...\right)+\left(...\right)\)
\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3.4A=3-1+\frac{1}{3}-...-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow12A+4A=\left(...\right)+\left(...\right)\)
\(\Rightarrow16A=3-\frac{101}{3^{99}}-\frac{100}{3^{100}}< 3\)
\(\Rightarrow A< \frac{3}{16}\)