so sanh
a)\(A=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+.....+\frac{5}{4^{99}}vaB=\frac{5}{3}\)
b)\(B=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+.....+\frac{3^{98}+1}{3^{98}}vaA=100\)
Những câu hỏi liên quan
Chứng minh rằng:
\(A=\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{n^3}< \frac{1}{4}\)
\(C=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98+1}}{3^{98}}< 100\)
\(D=\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{39}}< \frac{5}{3}\)
Bạn nào giải đúng mình tick đúng cho.Nhớ là phải giải nhanh đó
a, Ta có:
\(\frac{1}{2^3}< \frac{1}{1\cdot2\cdot3};\frac{1}{3^3}< \frac{1}{2\cdot3\cdot4};\frac{1}{4^3}< \frac{1}{3\cdot4\cdot5};...;\frac{1}{n^3}< \frac{1}{\left[n-1\right]n\left[n+1\right]}\)
\(\Rightarrow\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{3^3}+...+\frac{1}{n^3}< \frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left[n-1\right]n\left[n+1\right]}\)
Đặt \(A'=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left[n-1\right]n\left[n+1\right]}\)
\(\Rightarrow\frac{1}{2}A'=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{\left[n-1\right].n}-\frac{1}{n\left[n+1\right]}\)
\(\frac{1}{2}A'=\frac{1}{1\cdot2}-\frac{1}{n\left[n+1\right]}=\frac{1}{2}-\frac{1}{n\left[n+1\right]}=\frac{1}{4}-\frac{1}{2n\left[n+1\right]}< \frac{1}{4}\)
Vậy \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left[n-1\right]n\left[n+1\right]}< \frac{1}{4}\Leftrightarrow\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{n^3}< \frac{1}{4}\)
b,
\(C=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}=1+\frac{1}{3}+1+\frac{1}{3^2}+1+\frac{1}{3^3}+...+1+\frac{1}{3^{98}}\)
\(=\left[1+1+1+...+1\right]+\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right]=98+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
Đặt \(C'=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3C'=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{97}}\)
\(\Rightarrow3C'-C'=\left[1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\right]-\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right]=1-\frac{1}{3^{98}}\)
\(\Rightarrow C'=\frac{1-\frac{1}{3^{98}}}{2}< 1\)
\(\Rightarrow98+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}< 98+1=99< 100\)
\(\Rightarrow\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}< 100\)
c,
\(D=\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{39}}\)
\(4D=5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{38}}\)
\(4D-D=\left[5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{38}}\right]-\left[\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{38}}+\frac{5}{4^{39}}\right]\)
\(3D=5-\frac{5}{4^{39}}\Leftrightarrow D=\frac{5-\frac{5}{4^{39}}}{3}< \frac{5}{3}\)
Vậy:...........
AI THẤY ĐÚNG NHỚ ỦNG HỘ NHA
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Chứng minh rằng:
1)B=\(\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}< 100\)
2)C=\(\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)<\(\frac{1}{48}\)
3)D=\(\frac{11}{9}+\frac{18}{16}+\frac{27}{25}+...+\frac{1766}{1764}\)
\(40\frac{20}{43}< D< 40\frac{20}{21}\)
Chứng minh rằng
a) \(\frac{1}{5}<\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}<\frac{2}{5}\)
b) \(\frac{1}{15}<\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}<\frac{1}{10}\)
OK. Tối nhớ giải hộ mik nha
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\(\left(1\right)\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}>\frac{1}{5}\)
\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{13}{60}+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+...\left(\frac{1}{98}-\frac{1}{99}\right)\)
Ta thấy \(\frac{13}{60}>\frac{12}{60}=\frac{1}{5}\)
\(\frac{1}{6}-\frac{1}{7}>0\)
\(\frac{1}{8}-\frac{1}{9}>0\)
\(...\)\(\frac{1}{98}-\frac{1}{99}>0\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}>\frac{1}{5}\)
\(\left(2\right)\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}< \frac{2}{5}\)
\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)-\left(\frac{1}{9}-\frac{1}{10}\right)-...-\left(\frac{1}{97}-\frac{1}{98}\right)-\frac{1}{99}\)
\(=\frac{23}{60}-\left(\frac{1}{7}-\frac{1}{8}\right)-\left(\frac{1}{9}-\frac{1}{10}\right)-...-\left(\frac{1}{97}-\frac{1}{98}\right)-\frac{1}{99}\)
Ta thấy \(\frac{23}{60}< \frac{24}{60}=\frac{2}{5}\)
\(\frac{1}{7}-\frac{1}{8}>0\)
\(\frac{1}{9}-\frac{1}{10}>0\)
\(...\frac{1}{97}-\frac{1}{98}>0\)
\(\frac{1}{99}>0\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}< \frac{2}{5}\)
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1. Tính tích N=\(\frac{1}{2}.\frac{-2}{3}.\frac{3}{4}.\frac{-4}{5}.\frac{5}{6}.\frac{-6}{7}.......\frac{-98}{99}.\frac{99}{100}\)
\(\frac{1+2}{2+3}+\frac{2+3}{3+4}+\frac{3+4}{4+5}+...+\frac{98+99}{99+100}\)
Sửa dấu + thành dấu nhân
\(=\frac{2}{3}.\frac{5}{7}.\frac{7}{9}.....\frac{197}{199}=\frac{10}{597}\)
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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
= \(1+\frac{1}{3}+1+\frac{1}{9}+1+\frac{1}{27}+...+1+\frac{1}{3^{98}}\)\(\frac{1}{3^{98}}\)
\(=1.98+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
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.
A=\(\left(\frac{1}{4}+\frac{1}{^{4^2}}+\frac{1}{4^3}+\frac{1}{4^4}\right).4^5+\left(\frac{1}{4^5}+\frac{1}{4^6}+\frac{1}{4^7}+\frac{1}{4^8}\right).4^9+......+\left(\frac{1}{4^{97}}+\frac{1}{4^{98}}\frac{1}{4^{99}}+\frac{1}{4^{100}}\right).4^{101}\)
A=( 4^5/4+4^5/4^2+4^5/4^3+4^5/4^4 )+.....................+ ( 4^101/4^97+....+4^101/4^100 )
A = ( 4^4+ 4^3+4^2+4 ) + .........................................+ ( 4^4 + 4^3+4^2+4)
A= ( 4^4 + 4^ 3+ 4^2+4 ) * ( (101-5):4+1)
A = (4^4+4^3+4^2+4) * 25
A =( 256+81+16+4)*25= 8925
k cho mình nhé
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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
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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