Chứng minh rằng :
\(\frac{1}{3}\) - \(\frac{2}{3^2}\) + \(\frac{4}{3^4}\) +...............+\(\frac{99}{3^{99}}\) - \(\frac{100}{3^{100}}\) < \(\frac{3}{16}\)
Giải chi tiết giúp mk nha các bn!! mk cảm ơn nhìu ạ!!
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11/ chứng minh rằng:
b) \(\frac{1}{3}-\frac{2}{3^2} +\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)
Giải chi tiết mình like cho .
BẠN NÀO GIÚP MÌNH VỚI
CHỨNG MINH RẰNG:
\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
CÁC BẠN GIẢI ĐẦY ĐỦ HỘ MÌNH NHA
Đặt \(A=\frac{1}{3}-\frac{2}{3^2}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(4A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
Đặt \(B=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3B=3+1+...+\frac{3}{3^{98}}\)
\(2B=3-\frac{1}{3^{99}}\)
\(B=\frac{3}{2}-\frac{1}{3^{99}.2}\)
Thay B vào 4A ta có:
\(4A=\frac{3}{2}-\frac{1}{3^{99}.2}\)
\(A=\frac{3}{2.4}-\frac{1}{3^{99}.2.4}\)
\(A=\frac{3}{8}-\frac{1}{3^{99}.8}\)
Vì \(\frac{3}{8}>\frac{3}{16}\)
\(\Rightarrow\frac{3}{8}-\frac{1}{3^{99}.8}< \frac{3}{16}\)
Vậy \(A< \frac{3}{16}\)
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Chứng minh rằng:
a,\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b,\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
giúp minh với
Chứng minh rằng: a)\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b)\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Nhanh lên nhé! Mk đang cần gấp.
\(3B=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(B=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4B=3B+B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
+ Đặt \(M=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(3M=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(\Rightarrow4M=3M+M=3-\frac{1}{3^{99}}\)
\(\Rightarrow M=\frac{3}{4}-\frac{1}{3^{99}\cdot4}\)
\(\Rightarrow4B=M-\frac{100}{3^{100}}=\frac{3}{4}-\frac{1}{3^{99}\cdot4}-\frac{100}{3^{100}}\)
\(\Rightarrow B=\frac{3}{16}-\frac{1}{3^{99}\cdot16}-\frac{100}{3^{100}\cdot4}\) \(\Rightarrow B< \frac{3}{16}\)
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a) \(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
\(\Rightarrow3A=2A+A=1-\frac{1}{2^6}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{2^6\cdot3}< \frac{1}{3}\) ( đpcm )
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Chứng minh rằng \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\frac{5}{3^5}-.........+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Chứng minh rằng:
B = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)
3B=\(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+..........+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3B+B=\(1-\frac{1}{3}+\frac{1}{3^2}-..........+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
4B<\(1-\frac{1}{3}+\frac{1}{3^2}-.........+\frac{1}{3^{99}}\)
12B<\(3-1+\frac{1}{3}-.........+\frac{1}{3^{98}}\)
12B+4B<\(3-\frac{1}{3^{99}}\)
16B<3
\(\Rightarrow B<\frac{3}{16}\)
\(\Rightarrow\)B
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3B = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+.....-\frac{100}{3^{99}}\)
B + 3B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+.....+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
4B = M - \(\frac{100}{3^{100}}\) Với M = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+......+\frac{1}{3^{99}}\)
Ta lại có : 3M = 3 -1 +\(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-......+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
M + 3M = 3 - \(\frac{1}{3^{99}}\)
4M = 3 - 1/399 => M = 3/4 - 1/4.399
Khi đó : 4A = ( 3/4 - 1/4.399) - 1/399
4A = 3/4 - 1/4.399 - 1/399 < 3/4
=> A < 3/4 : 4
=> A < 3/16 (đpcm)
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Chứng minh rằng: \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
CÁC Bạn Làm Ơn Giải Hộ MK bài Này Với Ạ,mk đang cần rất gấp!!!!!!
Chứng Minh Rằng
a) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Chứng minh rằng : \(\frac{5}{6}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{11}{16}\)