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\(3C=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}\)
\(2C=3C-C=1-\frac{1}{3^{99}}\Rightarrow C=\left(1-\frac{1}{3^{99}}\right):2=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)
3C = 1+1/3+1/3^2+....+1/3^98
2C = 3C - C = (1+1/3+1/3^2+...+1/3^98) - (1/3+1/3^2+1/3^3+...+1/3^99) = 1- 1/3^99 < 1
=> C < 1/2
k mk nha
Câu hỏi của Ngô Văn Nam - Toán lớp 6 - Học toán với OnlineMath
Ta có :
M = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
3M = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
3M - M = ( \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)) - ( \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\))
2M = \(1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow M=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
3M=1+1/3+1/3^2+....+1/3^98
2M=3M-M=(1+1/3+1/3^2+....+1/3^98)-(1/3+1/3^2+....+1/3^99) = 1-1/3^99 < 1
=> M < 1/2
=> ĐPCM
k mk nha
b) A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
3A=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
3A-A=\(1-\frac{1}{3^{99}}\)
2A=\(1-\frac{1}{3^{99}}\)
vì 2A<1
=> A<\(\frac{1}{2}\)
\(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3C=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^2}+...+\frac{99}{3^{89}}-\frac{100}{3^{99}}\)
\(\Rightarrow4C=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4C< 1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\left(1\right)\)
Đặt: \(B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(\Rightarrow3B=2+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(4B=B+3B=3-\frac{1}{3^{99}}< 3\)
\(\Rightarrow B< \frac{3}{4}\left(2\right)\)
Từ: \(\left(1\right)\left(2\right)\Rightarrow4C< B< \frac{3}{4}\)
\(\Rightarrow C< \frac{3}{16}\left(đpcm\right)\)
(Đánh nhanh quá sai chỗ nào thông cảm nha :))
\(C=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
=> \(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
=> \(2C=1-\frac{1}{3^{99}}\)
=> \(C=\frac{1-\frac{1}{3^{99}}}{2}\)
Vì\(1-\frac{1}{3^{99}}< 1\Rightarrow\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
ngu vậy dat S roi tinh