B=\(\frac{4}{3}\)+\(\frac{10}{9}\)+\(\frac{28}{27}\)+...+\(\frac{3^{98}+1}{3^{98}}\)
CMR:B<100
Những câu hỏi liên quan
Cho biểu thức B = \(\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}.\).Chứng minh B <100
= \(1+\frac{1}{3}+1+\frac{1}{9}+1+\frac{1}{27}+...+1+\frac{1}{3^{98}}\)\(\frac{1}{3^{98}}\)
\(=1.98+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
Đặt A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)
\(\Rightarrow3A-A=2A=1-\frac{1}{3^{98}}\Rightarrow A=\frac{1-\frac{1}{2^{98}}}{2}< 1\)
\(\Rightarrow B=98+A< 98+1< 99< 100\)
\(\Rightarrow B< 100\)
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Cho biểu thức: B =\(\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...\frac{3^{98}+1}{3^{98}}\). Chứng minh B < 100.
cho biểu thức B=\(\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}\).chứng minh: B<100
Xét \(B=\frac{4}{3}+\frac{10}{9}+...+\frac{3^{98}+1}{3^{98}}\)
\(\Leftrightarrow B=\frac{3+1}{3}+\frac{9+1}{9}+...+\frac{3^{98}+1}{3^{98}}\)
\(\Leftrightarrow B=\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{1}{3^{98}}\right)\)(có 98 cặp số hạng)
\(\Leftrightarrow B=\left(1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)\)(có 98 số hạng 1)
\(\Leftrightarrow B=98+\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)\)
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)
Lấy 3A-A, ta được:
\(2A=1-\frac{1}{3^{98}}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2\cdot3^{98}}\)(*)
Thay (*) vào biểu thức B, ta được
\(B=98+\frac{1}{2}-\frac{1}{2\cdot3^{98}}< 100\)
VẬY, B<100 (ĐPCM)
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Ta có :
\(B=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}\)
\(B=\frac{3+1}{3}+\frac{9+1}{9}+\frac{27+1}{27}+...+\frac{3^{98}+1}{3^{98}}\)
\(B=\frac{3}{3}+\frac{1}{3}+\frac{9}{9}+\frac{1}{9}+\frac{27}{27}+\frac{1}{27}+...+\frac{3^{98}}{3^{98}}+\frac{1}{3^{98}}\)
\(B=1+\frac{1}{3}+1+\frac{1}{9}+1+\frac{1}{27}+...+1+\frac{1}{3^{98}}\)
\(B=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^{98}}\right)\)
\(B=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
Do từ \(1\) đến \(98\) có \(98-1+1=98\) số hạng nên có \(98\) số \(1\) suy ra :
\(B=98+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\) ta có :
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
\(2A=1-\frac{1}{3^{98}}< 1\)
Mà \(2A< 1\)\(\Rightarrow\)\(A< 1\)
Do đó :
\(B=98+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)< 98+1=99< 100\)
\(\Rightarrow\)\(B< 100\) ( đpcm )
Vậy \(B< 100\)
Chúc bạn học tốt ~
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Cho \(B=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}\)
Chứng minh B < 100
Cho biểu thức:B=\(\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+.....+\frac{3^{98}+1}{3^{98}}\).Chứng minh rằng B<100
B = \(\frac{4}{3^1}+\frac{10}{3^2}+\frac{28}{3^3}+...+\frac{3^{98}+1}{3^{98}}\)
B = \(\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{3^2}\right)+\left(1-\frac{1}{3^3}\right)+...+\left(1-\frac{1}{3^{98}}\right)\)
B = \(\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
B = \(98-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
=> B < 98 < 100
vậy B < 100
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\(B=\frac{4}{3}+\frac{10}{3^2}+...+\frac{3^{98}+1}{3^{98}}\)
\(3B=3\left(\frac{4}{3}+\frac{10}{3^2}+...+\frac{3^{98}+1}{3^{98}}\right)\)
\(3B=4+\frac{10}{3}+...+\frac{3^{98}+1}{3^{97}}\)
\(3B=\left(4+\frac{10}{3}+...+\frac{3^{98}+1}{3^{97}}\right)-\left(\frac{4}{3}+\frac{10}{3^2}+...+\frac{3^{98}+1}{3^{98}}\right)\)
\(2B=4-\frac{3^{98}+1}{3^{98}}\)
\(B=2-\frac{3^{98}+1}{2.3^{98}}<2\)
mà 2<100
=>B<100
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Xem thêm câu trả lời
Cho B = \(\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}\)
Chứng minh rằng B < 100
\(B=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{3^2}\right)+...+\left(1-\frac{1}{3^{98}}\right)\)
\(B=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)\)
\(B=98-\left(\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)<98\)
=>B<98<100
=>B<100 (đpcm)
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so sanh
a)\(A=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+.....+\frac{5}{4^{99}}vaB=\frac{5}{3}\)
b)\(B=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+.....+\frac{3^{98}+1}{3^{98}}vaA=100\)
\(B=\frac{4^3}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}-1}{3^{98}}\).\(CM:B<100\)
Cho : B=\(\frac{4}{3}\)+\(\frac{10}{9}\)+\(\frac{28}{27}\)+...+\(\frac{3^{98}+1}{3^{98}}\)CM : B<100
\(B=\frac{3^n+1}{3^n}=1+\frac{1}{3^n}=C+D\)
B có 98 số hạng => C=98
\(D=\frac{1}{3}+\frac{..1}{3^{97}}+\frac{1}{3^{98}}\)
3.D=1+1/3+....+1/3^97
tRỪ CHO NHAU
2D=1-1/3^98
\(C=\frac{1}{2}-\frac{1}{2.3^{98}}< \frac{1}{2}\)
\(B=98+\frac{1}{2}-\frac{1}{2.3^{98}}< 99< 100\) có lẽ đề lấy 100 co chẵn. hay cộng nhầm ai tets hộ cái
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