D = \(\frac{1}{3}\) + \(\frac{2}{3^2}\)+ \(\frac{3}{3^3}\)+ .... + \(\frac{101}{3^{101}}\)
CMR : D < \(\frac{3}{4}\)
Những câu hỏi liên quan
D = \(\frac{1}{3}\)+ \(\frac{2}{3^2}\)+\(\frac{3}{3^3}\)+.... + \(\frac{101}{3^{101}}\)
CMR : D < \(\frac{3}{4}\)
26 tháng 7 2015 lúc 13:03
CMR: \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{101}{3^{101}}+\frac{102}{3^{102}}<\frac{3}{4}\)
mik biet lam nhung ngại viet lam ban thong cam nha
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Cho biểu thức D =\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^2}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\) chứng minh rằng D < \(\frac{3}{4}\)
D=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^2}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)
D=\(\frac{1}{3}+\frac{101}{3^{101}}\)
D=\(\frac{1}{3}\)
\(\frac{1}{3}và\frac{3}{4}\)
\(\frac{1}{3}=\frac{4}{12}\)
\(\frac{3}{4}=\frac{9}{12}\)
Vì\(\frac{4}{12}< \frac{9}{12}Vậy\frac{1}{3}< \frac{3}{4}\)
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Cho biểu thức: D= \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)
Chứng minh rằng D < \(\frac{3}{4}\)
CMR:
a)\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\) <1
b)\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{4^4}+....+\frac{101}{3^{101}}\),<3/4
nhanh nhé
a) Ta có
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^6}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)
\(A=1-\frac{1}{2^7}\)
Do \(1-\frac{1}{2^7}< 1\Rightarrow A< 1\left(đpcm\right)\)
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C = \(\frac{1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}}{1+\frac{1}{3}+......+\frac{1}{3^{101}}}\)
D = \(\frac{1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{101}}}{1+\frac{1}{3}+........+\frac{1}{3^{102}}}\)
ai lam ho minh voi so sanh nha
cho biểu thức D = \(\frac{1}{3}\)+ \(\frac{2}{3^2}\)+ \(\frac{3}{3^3}\)+ ......+ \(\frac{100}{3^{100}}\)+ \(\frac{101}{3^{101}}\)
Chứng minh rằng D < \(\frac{3}{4}\)
A1+frac{3}{2^3}+frac{4}{2^4}+...+frac{100}{2^{100}}frac{A}{2}frac{1}{2}+frac{3}{2^4}+frac{4}{2^5}+....+frac{100}{2^{101}}A-frac{A}{2}left(1+frac{3}{2^3}+....+frac{100}{2^{100}}right)-left(frac{1}{2}+frac{3}{2^4}+.....+frac{100}{2^{101}}right)frac{A}{2}frac{1}{2}+frac{3}{2^3}+frac{1}{2^4}+frac{1}{2^5}+....+frac{1}{2^{100}}-frac{100}{2^{101}}frac{A}{2}frac{1}{2}+frac{1}{2^2}+frac{1}{2^3}+frac{1}{2^4}+....+frac{1}{2^{100}}-frac{1}{2^{101}}frac{A}{2}left(1-left(frac{1}{2}right)^{101}right).2-frac{10...
Đọc tiếp
\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+....+\frac{100}{2^{101}}\)\(A-\frac{A}{2}=\left(1+\frac{3}{2^3}+....+\frac{100}{2^{100}}\right)-\left(\frac{1}{2}+\frac{3}{2^4}+.....+\frac{100}{2^{101}}\right)\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+....+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^{100}}-\frac{1}{2^{101}}\)
\(\frac{A}{2}=\left(1-\left(\frac{1}{2}\right)^{101}\right).2-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{2^{101}-1}{2^{100}}-\frac{100}{2^{101}}\)
\(A=\frac{2^{101}-1}{2^{99}}-\frac{100}{2^{100}}\)
Chứng minh:
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{101}{3^{101}}