Cho \(A=\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^n-1}{+3^n}\)
\(CMR:A>n-\frac{1}{2}\)
Những câu hỏi liên quan
Cho A = \(\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+......+\frac{3^n-1}{3^n}\) CMR A > n-\(\frac{1}{2}\)
\(A=\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^n-1}{3^n}\)
\(=\frac{3-1}{3}+\frac{9-1}{9}+\frac{27-1}{27}+...+\frac{3^n-1}{3^n}\)
\(=\left(\frac{3}{3}-\frac{1}{3}\right)+\left(\frac{9}{9}-\frac{1}{9}\right)+\left(\frac{27}{27}-\frac{1}{27}\right)+.....+\left(\frac{3^n}{3^n}-\frac{1}{3^n}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....+\frac{1}{3^n}\right)\)
\(=n-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^n}\right)\)
Bây giờ ta chỉ cần chứng minh:\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^n}< \frac{1}{2}\) là xong!
Thật vậy:\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{n-1}}\)
\(\Rightarrow2B=1-\frac{1}{3^n}\)
\(\Rightarrow B=\frac{1}{2}-\frac{\frac{1}{3^n}}{2}< \frac{1}{2}\)
Ta có:\(A=n-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^n}\right)\)
\(>n-\frac{1}{2}\left(đpcm\right)\)(bất đẳng thức đổi chiều)
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\(choA=\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^n-1}{3^n}\)
Chứng minh rằng \(A< n-\frac{1}{2}\)
18 tháng 10 2015 lúc 15:25
Cho \(A=\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^n-1}{3^n}.\)
Chứng minh : \(A>n-\frac{1}{2}\)
A =\(\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+.....+\frac{3^n-1}{3^n}\). Chứng minh rằng A > n - \(\frac{1}{2}\)
1 CMR:
B=\(\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+.....+\frac{3n+1}{3^n}< \frac{11}{4}\)(n thuộc N*;n>3)
A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
C=\(\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^{20}-1}{3^{20}}>19\frac{1}{2}\)
Có : \(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3A-A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow2A< 1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
Có: \(6A< 3+1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(6A-2A< 3-\frac{1}{3^{99}}< 3\)
\(\Rightarrow4A< 3\Rightarrow A< \frac{3}{4}\)(đpcm)
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CHO \(A=\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3n-1}{3n}\).\(CM:A>n-\frac{1}{2}\)
trang 61 SGK Toán 11 tập 1 - Chân trời sáng tạo
17 tháng 7 2023 lúc 12:36
Cho dãy số \(\left( {{u_n}} \right)\) với \({u_n} = \frac{n}{{{3^n} - 1}}\). Ba số hạng đầu tiên của dãy số \(\left( {{u_n}} \right)\) lần lượt là:
A. \(\frac{1}{2};\frac{1}{4};\frac{3}{{27}}\).
B. \(\frac{1}{2};\frac{1}{4};\frac{3}{{26}}\).
C. \(\frac{1}{2};\frac{1}{4};\frac{3}{{25}}\).
D. \(\frac{1}{2};\frac{1}{4};\frac{3}{{28}}\).
Ta có:
\(u_1=\dfrac{1}{3^1-1}=\dfrac{1}{2}\\ u_2=\dfrac{2}{3^2-1}=\dfrac{1}{4}\\ u_3=\dfrac{3}{3^3-1}=\dfrac{3}{26}\)
\(\Rightarrow B\)
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\(CMR:A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{n}\left(n\in N;n\ge2\right)\)
Tinh :
M = \(\frac{\frac{3}{20}-\frac{1}{15}+\frac{1}{195}}{\frac{1}{2}-\frac{1}{26}-\frac{9}{8}}\): 2,1(3)
N = \(\frac{1!}{4!}+\frac{2!}{5!}+\frac{3!}{6!}+...+\frac{12!}{15!}\)