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\(\text{Đặt biểu thức là A:}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}\)
\(\text{Ta có:}\frac{1}{2^2}=\frac{1}{2\times2}< \frac{1}{1\times2}\)
\(\frac{1}{3^2}=\frac{1}{3\times3}< \frac{1}{2\times3}\)
\(\frac{1}{4^2}=\frac{1}{4\times4}< \frac{1}{3\times4}\)
\(...\)
\(\frac{1}{99^2}=\frac{1}{99\times99}< \frac{1}{98\times99}\)
\(\frac{1}{100^2}=\frac{1}{100\times100}=\frac{1}{99\times100}\)
\(\Rightarrow A< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).......\left(\frac{1}{100^2}-1\right)\)
\(A=\left(\frac{1}{2^2}-\frac{2^2}{2^2}\right).\left(\frac{1}{3^2}-\frac{3^2}{3^2}\right).....\left(\frac{1}{100^2}-\frac{100^2}{100^2}\right)\)
\(A=\left(-\frac{3}{4}\right).\left(-\frac{8}{9}\right)........\left(-\frac{9999}{10000}\right)\)
\(A=\frac{\left(-3\right).\left(-8\right).....\left(-9999\right)}{4.9...10000}=\frac{1.\left(-3\right).2.\left(-4\right)......99.\left(-101\right)}{2.2.3.3.....100.100}\)
\(A=\frac{\left(1.2.3....99\right).\left[\left(-3\right).\left(-4\right)......\left(-101\right)\right]}{\left(2.3.4....100\right).\left(2.3.4...100\right)}=\frac{1.\left(-101\right)}{100.\left(-1.\right).\left(-1\right)....\left(-1\right).2}=\frac{-101}{100.2}=\frac{-101}{200}\)
Ta thấy \(\frac{-101}{200}< \frac{-100}{200}=\frac{-1}{2}\Rightarrow A< -\frac{1}{2}\)
P \(=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
P\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{50^2-1}{50^2}\)
P \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
P\(=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
P\(=\frac{1.51}{50.2}=\frac{51}{100}\)
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 \(C=\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{100}}\)
\(\Leftrightarrow2C=1+\frac{1}{2}+.......+\frac{1}{2^{99}}\)
\(\Leftrightarrow2C-C=\left(1+\frac{1}{2}+.....+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+.....+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow C=1-\frac{1}{2^{100}}>1\)
\(\Rightarrow A>B\)
Ta có :
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(A=\left(\frac{1}{2^2}-\frac{2^2}{2^2}\right).\left(\frac{1}{3^2}-\frac{3^2}{3^2}\right).\left(\frac{1}{4^2}-\frac{4^2}{4^2}\right)...\left(\frac{1}{100^2}-\frac{100^2}{100^2}\right)\)
\(A=\left(-\frac{3}{2^2}\right).\left(-\frac{8}{3^2}\right).\left(-\frac{15}{4^2}\right)...\left(-\frac{99}{100^2}\right)\)
\(A=-\left(\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4....10.10}\right)\)
\(A=-\left(\frac{1.2.3....9}{2.3.4....10}.\frac{3.4.5.....11}{2.3.4....10}\right)\)
\(A=-\left(\frac{1}{10}.\frac{11}{2}\right)=-\frac{11}{20}=\frac{-11}{20}\)
Lại có : \(\frac{-1}{2}=\frac{-1.10}{2.10}=\frac{-10}{20}\)
Vì \(-11< -10\)nên \(\frac{-11}{20}< \frac{-10}{20}\)hay \(A< \frac{-1}{2}\)
Mk mới học bài này xong,nhớ ủng hộ mk nha !!! ^_^
A=(\(\frac{1}{2^2}\) -1).( \(\frac{1}{3^2}\)-1)............(\(\frac{1}{100^2}\) -1)=\(-\frac{\left(1.2.3.4....99\right)\left(1.2.3.4....101\right)}{\left(1.2.3.4....100\right)\left(1.2.3.4....100\right)}\)=\(\frac{-101}{100}\)
Ta có :
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(A=1-\frac{1}{2^{99}}< 1\\ \)