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BÀI 1:
\(P=1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{2^{100}-1}\)
\(\Leftrightarrow A=1+\frac{1}{2}+\frac{1}{3}+..........+\frac{1}{2^{100}-1}+\frac{1}{2^{100}}-\frac{1}{2^{100}}\)
\(\Leftrightarrow A=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{2^2}\right)+........+\left(\frac{1}{2^{99}+1}+.......+\frac{1}{2^{100}}\right)-\frac{1}{2^{100}}\)
\(\Leftrightarrow A>1+\frac{1}{2}+\frac{1}{2^2}\cdot2+\frac{1}{2^3}\cdot2^2+........+\frac{1}{2^{100}}\cdot2^{99}-\frac{1}{2^{100}}\)
\(\Leftrightarrow A>1+\frac{1}{2}\cdot100-\frac{1}{2^{100}}\)
\(\Leftrightarrow A>51-\frac{1}{2^{100}}>51-1=50\)
\(\Rightarrow DPCM\)
BÀI 2 :
TA CÓ: \(A=1+\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{100}}\)VÀ \(B=2\)
= > CẦN CHỨNG MINH \(\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{100}}\)NHƯ THẾ NÀO SO VỚI 1
ĐẶT \(C=\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{100}}\)
\(\Leftrightarrow2C=1+\frac{1}{2}+.......+\frac{1}{2^{99}}\)
\(\Leftrightarrow2C-C=\left(1+\frac{1}{2}+.....+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+.....+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow C=1-\frac{1}{2^{100}}>1\)
\(\Rightarrow A>B\)
2 vế bằng nhau
100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100
100- 1-1/2-1/3-...-1/100 = 1/2+2/3+3/4+...+99/100
100 = 1 + 1/2 + 1/2 + 1/3 + 2/3 + ... + 1/100 + 99/100 (cùng cộng 2 vế với (- 1-1/2-1/3-...-1/100)
100 = 1 + 1 + 1 + ... + 1 (100 số hạng)
100 = 100
Vậy 100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100
A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
A < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
A < 1 - \(\frac{1.}{100}\)
A < \(\frac{99}{100}< \frac{199}{100}\)
=> A < \(\frac{199}{100}\)
b,
S = \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{99}{10^2}\)
S = \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{9.11}{10.10}\)
S = \(\frac{1.3.2.4.3.5.4.6.5.7...9.11}{2.2.3.3.4.4...10.10}\)
S = \(\frac{1.2.3^2.4^2.5^2...9^2.10.11}{2^2.3^3.4^2...10^2}\)
S = \(\frac{1.11}{2.10}\)
S = \(\frac{11}{20}\)
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).......\left(\frac{1}{100^2}-1\right)\)
\(A=\left(\frac{1}{2^2}-\frac{2^2}{2^2}\right).\left(\frac{1}{3^2}-\frac{3^2}{3^2}\right).....\left(\frac{1}{100^2}-\frac{100^2}{100^2}\right)\)
\(A=\left(-\frac{3}{4}\right).\left(-\frac{8}{9}\right)........\left(-\frac{9999}{10000}\right)\)
\(A=\frac{\left(-3\right).\left(-8\right).....\left(-9999\right)}{4.9...10000}=\frac{1.\left(-3\right).2.\left(-4\right)......99.\left(-101\right)}{2.2.3.3.....100.100}\)
\(A=\frac{\left(1.2.3....99\right).\left[\left(-3\right).\left(-4\right)......\left(-101\right)\right]}{\left(2.3.4....100\right).\left(2.3.4...100\right)}=\frac{1.\left(-101\right)}{100.\left(-1.\right).\left(-1\right)....\left(-1\right).2}=\frac{-101}{100.2}=\frac{-101}{200}\)
Ta thấy \(\frac{-101}{200}< \frac{-100}{200}=\frac{-1}{2}\Rightarrow A< -\frac{1}{2}\)
\(\text{Đặt biểu thức là A:}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}\)
\(\text{Ta có:}\frac{1}{2^2}=\frac{1}{2\times2}< \frac{1}{1\times2}\)
\(\frac{1}{3^2}=\frac{1}{3\times3}< \frac{1}{2\times3}\)
\(\frac{1}{4^2}=\frac{1}{4\times4}< \frac{1}{3\times4}\)
\(...\)
\(\frac{1}{99^2}=\frac{1}{99\times99}< \frac{1}{98\times99}\)
\(\frac{1}{100^2}=\frac{1}{100\times100}=\frac{1}{99\times100}\)
\(\Rightarrow A< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
=>\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
=>\(A=1-\frac{1}{2^{100}}< 1\)
Vậy A<1