Cho \(A=1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^3+\left(\frac{3}{4}\right)^4-...-\left(\frac{3}{4}\right)^{2009}+\left(\frac{3}{4}\right)^{2010}\)
Chứng tỏ A ko phải là số nguyên
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\(A=1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^4-...-\left(\frac{3}{4}\right)^{2009}+\left(\frac{3}{4}\right)^{2010}\)
\(\Rightarrow\frac{3}{4}A=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^2+-\left(\frac{3}{4}\right)^4+...+\left(\frac{3}{4}\right)^{2010}-\left(\frac{3}{4}\right)^{2011}\)
\(\Rightarrow\frac{3}{4}A+A=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^2+-\left(\frac{3}{4}\right)^4+...+\left(\frac{3}{4}\right)^{20010}-\left(\frac{3}{4}\right)^{2011}\)
\(+1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^4-...-\left(\frac{3}{4}\right)^{2009}+\left(\frac{3}{4}\right)^{2010}\)
\(\Rightarrow\frac{7}{4}A=1-\left(\frac{3}{4}\right)^{2011}\)
\(\Rightarrow A=\frac{4}{7}-\frac{4}{7}.\left(\frac{3}{4}\right)^{2011}\)
\(\Rightarrow A=\frac{4}{7}-\frac{3^{2011}}{7.4^{2010}}\)
Vậy A không là số tự nhiên
\(\frac{3}{4}A=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3-\left(\frac{3}{4}\right)^4+\left(\frac{3}{4}\right)^5-....-\left(\frac{3}{4}\right)^{2010}\)
\(A+\frac{3}{4}A=1-\left(\frac{3}{4}\right)^{2010}\)
\(\frac{7}{4}A=1-\left(\frac{3}{4}\right)^{2010}\)
\(A=\frac{4}{7}\left(1-\left(\frac{3}{4}\right)^{2010}\right)khong\:làsốnguyên\)
\(A+\frac{3}{4}A=1+\left(\frac{3}{4}\right)^{2011}\)
\(\Leftrightarrow\frac{7}{4}A=1+\left(\frac{3}{4}\right)^{2011}\)
\(\Leftrightarrow A=\left(1+\left(\frac{3}{4}\right)^{2011}\right):\frac{7}{4}=\frac{4}{7}\left(1+\left(\frac{3}{4}\right)^{2011}\right)\)
Vì \(1<1+\left(\frac{3}{4}\right)^{2011}<1+\frac{3}{4}=\frac{7}{4}\)
=> 4/7 < A < 4/7 .7/4 =1 => A không là số nguyên
\(A=1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^3+...+\left(\frac{3}{4}\right)^{2010}\)
\(\frac{3}{4}A=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3-...-\left(\frac{3}{4}\right)^{2011}\)
\(\frac{3}{4}A-1=-\left[1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^3+...+\left(\frac{3}{4}\right)^{2010}\right]-\left(\frac{3}{4}\right)^{2011}\)
\(\frac{3}{4}A-1=A-\left(\frac{3}{4}\right)^{2011}\)
\(\frac{3}{4}A-A=-\left(\frac{3}{4}\right)^{2011}+1\)
\(-\frac{1}{4}A=1-\left(\frac{3}{4}\right)^{2011}\)
\(A=\frac{1-\left(\frac{3}{4}\right)^{2011}}{-\frac{1}{4}}=1:-\frac{1}{4}-\left(\frac{3}{4}\right)^{2011}:\left(-\frac{1}{4}\right)=-4+3\cdot\left(\frac{3}{4}\right)^{2010}\)
=>A không phải là số nguyên