Tính và so sánh: \(S=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}...+\frac{99}{49^2.50^2}\)\(T=\frac{1}{2^2-1^2}+\frac{1}{3^2-1^2}+\frac{1}{4^2-1^2}+...+\frac{1}{50^2-1^2}\)
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so sánh: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\) và 1
CMR:
a) \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}< \frac{1}{2}\)
b) \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)
a)Xét vế trái , ta có :
Gọi tổng các số hạng ở vế trái là A
=> A= \(\frac{1}{3}\)+\(\frac{1}{3^2}\)+ ... +\(\frac{1}{3^{99}}\)
=>3A = 1 + \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+ ... + \(\frac{1}{3^{98}}\)
=> 3A - A = 1 + \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+ ... + \(\frac{1}{3^{98}}\)- ( \(\frac{1}{3}\)+\(\frac{1}{3^2}\)+ ... +\(\frac{1}{3^{99}}\))
=> 2A = 1 - \(\frac{1}{3^{99}}\)
=> A = \(\frac{1}{2}\)- \(\frac{1}{3^{99}.2}\) < \(\frac{1}{2}\)
b)\(\frac{3}{1^2.2^2}\)+ \(\frac{5}{2^2.3^2}\)+ ... + \(\frac{19}{9^2.10^2}\)
= \(\frac{3}{1.4}\)+ \(\frac{5}{4.9}\)+ .... + \(\frac{19}{81.100}\)
= 1 - \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{9}\)+ ... + \(\frac{1}{81}\)- \(\frac{1}{100}\)
= 1 - \(\frac{1}{100}\) <1
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a,
\(\sum\limits^{99}_{x=1}\left(\frac{1}{3^x}\right)=\frac{1}{2}\)
bài a nó có ............
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Cho \(M=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{199}{99^2.100^2}\). So sánh \(M\) với \(1\)
Ta có
\(M=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^3.3^2}+.....+\frac{100^2-99^2}{99^2.100^2}\)
\(M=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+......+\frac{1}{99^2}-\frac{1}{100^2}\)
\(M=1-\frac{1}{100^2}< 1\)
=> M<1
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Cho M=\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{199}{99^2.100^2}\) . So sánh M với 1.
\(M=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{100^2-99^2}{99^2.100^2}\)
\(M=\frac{2^2}{1^2.2^2}-\frac{1^2}{1^2.2^2}+\frac{3^2}{2^2.3^2}-\frac{2^2}{2^2.3^2}+\frac{4^2}{3^2.4^2}-\frac{3^2}{3^2.4^2}+...+\frac{100^2}{99^2.100^2}-\frac{99^2}{99^2.100^2}\)
\(M=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=1-\frac{1}{100^2}
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Tính :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{199}{99^2.100^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=\frac{9999}{10000}\)
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\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=\frac{9999}{10000}\)
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Bài 1: So sánh:
a) 5^255 và 2^512 ( 2 cách)
b) 8^12 và 12^8
Bài 2: Chứng minh rằng:
a) A = \(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{99}<1\)
b) B = \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}<1\)
c) C = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}<\frac{3}{4}\)
Chứng minh rằng ;
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{199}{99^2.100^2}\) nhỏ hơn 1
Cho C = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)chứng minh rằng C < \(\frac{1}{2}\)CMR : \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)CMR : \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
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#)Giải :
Bài 1 :
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\Leftrightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(\Leftrightarrow3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(\Leftrightarrow2C=1-\frac{1}{3^{100}}\Leftrightarrow C=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1}{2}\Rightarrow C< \frac{1}{2}\left(đpcm\right)\)
Bài 2 :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{16}\right)+...+\left(\frac{1}{81}-\frac{1}{100}\right)=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
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So sánh
F = \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+......+\frac{19}{9^2.10^2}\)với 1
E = \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\)với \(\frac{1}{2}\)
ta có:
\(\frac{1}{11}\)>\(\frac{10}{20}\)
\(\frac{1}{12}\)>\(\frac{10}{20}\)
\(\frac{1}{13}\)>\(\frac{10}{20}\)
....
\(\frac{1}{19}\)>\(\frac{10}{20}\)
=>E >\(\frac{10}{20}\)
vậy E > \(\frac{1}{2}\)
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