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Ta có :
\(A=\frac{101}{1}+\frac{100}{2}+\frac{99}{3}+...+\frac{1}{101}\)
\(A=\left(101-1-...-1\right)+\left(\frac{100}{2}+1\right)+\left(\frac{99}{3}+1\right)+...+\left(\frac{1}{101}+1\right)\)
\(A=\frac{102}{102}+\frac{102}{2}+\frac{102}{3}+...+\frac{102}{101}\)
\(A=102\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}+\frac{1}{102}\right)\)
\(\Rightarrow\)\(\frac{A}{B}=\frac{102\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}=\frac{102}{1}=102\)
Vậy \(\frac{A}{B}=102\)
Chúc bạn học tốt ~
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(\Leftrightarrow3A=1+\frac{1}{3}+\frac{1}{3^{^2}}+...+\frac{1}{3^{98}}\)
\(\Leftrightarrow3A-A=1-\frac{1}{3^{99}}\)
\(\Leftrightarrow2A=1-\frac{1}{3^{99}}\)
\(\Leftrightarrow A=\left(1-\frac{1}{3^{99}}\right)\div2\)
Đặt A=1/3-2/3^2+3/3^3-4/3^4+...+99/3^99+100/3^100
=>A<1/16
3A=1-2/3+3/3^2-4/3^3+...+99/3^98+100/3^99
=>3A-A=(1-2/3+3/3^2-4/3^3+...+99/3^98+100/3^99)-(1/3-2/3^2+3/3^3-4/3^4+...+99/3^99+100/3^100)
2A=5/3^2-7/3^3+1/3^99-100/3^100
2A=1/3^2(5-7/3+1/3^97-100/3^98)
A=1/18.(8/3+1/3^97-100/3^98)
A=1/54.(8+1/3^96-100/3^97)
Vì 1/54<1/16
=>A<1/16(đpcm)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(\Rightarrow\)\(3A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(\Rightarrow\)\(3A-A=\)\(\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(\Rightarrow\)\(2A=3-\frac{1}{3^{100}}\)
\(\Rightarrow\)A = \(\frac{3-\frac{1}{3^{100}}}{2}\)