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=> 3E =1+2/3+3/3^2+...+100/3^99
=> 3E-E=1+1/3+1/3^2+...+1/3^99-100/3^100
=> 2E=1+1/3+1/3^2+...+1/3^99-100/3^100
=> 6E=3+1+1/3+1/3^2+....+1/3^98-100/3^99
=> 6E-2E=3-100/3^99+100/3^100
=> 4E=3-100/3^99+100/3^100
=> E=3/4 -100/3^99.4+100/3^100.4<3/4
Vậy E< 3/4
Đặt \(A=\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}\)
\(A=\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+\left(\frac{3}{97}+1\right)+...+\left(\frac{98}{2}+1\right)+1\) ( 99/1 = 99, tất cả 98 ( không tính 99/1) hạng tử trong A đều cộng với 1 , dư ra 1 chỗ cuối)
\(A=\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}+\frac{100}{100}\) ( 100/100=1)
\(A=100.\left(\frac{1}{2}+...+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}\right)\)
Thay A vào E, có:
\(E=\frac{100.\left(\frac{1}{2}+...+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(E=100\)
\(\Rightarrow E=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+....+\frac{98}{2}+1+1+...+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\) ( Có 99 số 1)
\(\Rightarrow\frac{\frac{1}{99}+1+\frac{2}{98}+\frac{3}{97}+1+...+\frac{98}{2}+1+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)(Nhóm 98 số 1 với 98 phân số đầu ở trên tử)mik viết thiếu nha sorry *-*
\(\Rightarrow E=\frac{\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}+\frac{100}{100}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(\Rightarrow E=\frac{\frac{100}{2}+\frac{100}{3}+\frac{100}{4}+...+\frac{100}{99}+\frac{100}{100}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(\Rightarrow E=\frac{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(\Rightarrow E=\frac{100.1}{1}=100\)
~Chúc bạn hok tốt~
\(E=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3E=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3E-E=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2E=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6E=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6E-2E=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4E=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4E=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4E=3-\frac{203}{3^{100}}< 3\)
=> 4E < 3 => E < 3/4
\(F=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(F=\left(\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)
\(F=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)-2.\left(\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)
\(F=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{50}}\right)\)
\(F=\frac{1}{2^{51}}+\frac{1}{2^{52}}+...+\frac{1}{2^{100}}\)
\(E=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2E=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2E-E=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(E=1-\frac{1}{2^{100}}\)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~