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\(A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(\Rightarrow\)\(3A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(\Rightarrow\)\(3A-A=\)\(\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(\Rightarrow\)\(2A=3-\frac{1}{3^{100}}\)
\(\Rightarrow\)A = \(\frac{3-\frac{1}{3^{100}}}{2}\)
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=1/2+1/2^2+...+1/2^100
2A=1+1/2+...+1/2^99
2A-A=1+[1/2+(-1/2)]+...+(1/2^99-1/2^99)-1/2^100
2A-A=1+0+...+0-1/2^100
A=1-1/2^100
A=1
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 2\left(đpcm\right)\)
Ta có : A = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\left(1\right)\)
=> 3A = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\left(2\right)\)
Lấy (2) trừ (1) theo vế ta có : \(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
=> \(2A=1-\frac{1}{3^{100}}\Rightarrow A=\left(1-\frac{3}{3^{100}}\right):2=\frac{1}{2}-\frac{3}{2.3^{100}}\)