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\(\frac{2}{3}+\frac{1}{3}=\frac{6+3}{3}=\frac{9}{3}=3\)
\(\frac{3}{4}+\frac{2}{4}+\frac{1}{4}=\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{1}{2}=1+\frac{1}{2}=1\frac{1}{2}=\frac{3}{2}\)
\(\frac{4}{5}+\frac{3}{5}+\frac{2}{5}+\frac{1}{5}=\left(\frac{4}{5}+\frac{1}{5}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)=2+2=4\)
\(\frac{5}{6}+\frac{4}{6}+\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\left(\frac{5}{6}+\frac{1}{6}\right)+\left(\frac{4}{6}+\frac{2}{6}\right)+\frac{1}{2}=1+1\)\(+\frac{1}{2}=2\frac{1}{2}=\frac{5}{2}\)
ngu LÊ MĨ LINH
theo thứ tự :1,6/4 =1 và 1/2,2,5/2,500
MỚI LÀM LÚC TỐI,HÊN QUÁ:
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(4A=3-\left(\frac{101}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{203}{3^{100}}\)
\(A=\frac{3}{4}-\frac{203}{3^{100}\cdot4}< \frac{3}{4}\)
#)Giải :
\(A=\frac{1}{3^1}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(A=\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\)
\(\Rightarrow2A=1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\)
\(\Rightarrow2A-A=A=\left(1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\right)-\left(\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\right)\)
\(\Rightarrow A=1+\frac{2}{9}-\frac{2}{9^{50}}=\frac{11}{9}-\frac{2}{9^{50}}\)
Có lẽ đúng .........................
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
Đặt A= \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=1-\frac{1}{3^{100}}\)
Hay \(2A=1-\frac{1}{3^{100}}\)
\(\Rightarrow A=\left(1-\frac{1}{3^{100}}\right):2\)
_Học tốt nha_
\(\frac{1}{2}S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{101}}\)
=> \(\frac{1}{2}S-S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}+\frac{1}{2^{101}}-\frac{1}{2^{100}}-...-\frac{1}{2}-1\)
<=> \(\frac{-1}{2}S=\frac{1}{2^{101}}-1\)
<=> \(S=2-\frac{1}{2^{100}}\)
Ta có :
S = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\left(1\right)\)
\(\Rightarrow2S=2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\left(2\right)\)
Lấy (2) - (1) ta được :
\(S=2-\frac{1}{2^{100}}=\frac{2^{101}-1}{2^{100}}\)
đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow A+3A=\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)+\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)\)
\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)<\(B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(\Rightarrow3B=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(\Rightarrow B+3B=\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)+\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)\)
\(\Rightarrow4B=3-\frac{1}{3^{98}}<3\)
\(\Rightarrow B<\frac{3}{4}\Rightarrow4A<\frac{3}{4}\Rightarrow A<\frac{3}{16}\)
\(\RightarrowĐPCM\)
a) Ta có
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^6}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)
\(A=1-\frac{1}{2^7}\)
Do \(1-\frac{1}{2^7}< 1\Rightarrow A< 1\left(đpcm\right)\)
3A = 1 + 1/3 + 1/3^2 + ... + 1/3^199
3A - A = ( 1 + 1/3 + 1/3^2 + ... + 1/3^99 ) - ( 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^100 )
2A = 1 - 1/3^100
A = ( 1 - 1/3^100 ) / 2
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(A=\frac{3^{100}-1}{3^{100}.2}\)
mk chỉ làm được đến đây thôi