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7 tháng 4 2019
Ta có:\(8^2\equiv1\left(mod9\right)\)
\(\Rightarrow\left(8^2\right)^{50}=8^{100}\equiv1\left(mod9\right)\)
\(\Rightarrow\left(8^{100}-1\right)⋮9\left(đpcm\right)\)
7 tháng 4 2019
Ta có 8\(\equiv\)-1(mod 9)=> 8100\(\equiv\)(-1)100\(\equiv\)1(mod 9)
=>8100-1\(⋮\)9(đpcm)
15 tháng 5 2016
\(P>\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdot\cdot\cdot\frac{99}{100}\cdot\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{98}{99}\right)\)
\(P>\frac{49}{50}>\frac{1}{15}\)
\(P^2<\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\frac{8}{9}\cdot....\cdot\frac{100}{101}\right)\)
\(P^2<\frac{1}{101}<\frac{1}{10}\)
\(\Rightarrow\frac{1}{15}
28 tháng 4 2018
ta có: \(A=\frac{1}{3}+\frac{2}{3^2}+...+\frac{100}{3^{100}}\)
\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}+\frac{2}{3^3}+...+\frac{100}{3^{101}}\)
\(\Rightarrow A-\frac{1}{3}A=\frac{1}{3}+\left(\frac{2}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{100}{3^{100}}-\frac{99}{3^{100}}\right)-\frac{100}{3^{101}}\)
\(\frac{2}{3}A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{100}{3^{101}}\)
+) Xét \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(\Rightarrow\frac{1}{3}B=\frac{1}{3^2}+\frac{1}{3^3}...+\frac{1}{3^{101}}\)
\(\Rightarrow B-\frac{1}{3}B=\frac{1}{3}-\frac{1}{3^{101}}\)
\(\frac{2}{3}B=\frac{1}{3}-\frac{1}{3^{101}}\)
\(\Rightarrow B=\left(\frac{1}{3}-\frac{1}{3^{101}}\right):\frac{2}{3}=\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{2}\)
Thay B vào A, ta có:
\(\frac{2}{3}A=\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{2}-\frac{100}{3^{101}}\)
\(\Rightarrow A=\left(\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{2}-\frac{100}{3^{101}}\right):\frac{2}{3}\)
\(A=\left(\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{2}-\frac{100}{3^{101}}\right).\frac{3}{2}\)
\(A=\frac{9}{4}.\left(\frac{1}{3}-\frac{1}{3^{101}}\right)-\frac{100.3}{3^{101}.2}=\frac{9}{4}.\left(\frac{1}{3}-\frac{1}{3^{101}}\right)-\frac{150}{3^{101}}\)
\(A=\frac{3}{4}-\frac{9}{4}.\frac{1}{3^{101}}-\frac{150}{3^{101}}< \frac{3}{4}\)
\(\Rightarrow A=\frac{1}{3}+\frac{2}{3^2}+...+\frac{100}{3^{100}}< \frac{3}{4}\left(đpcm\right)\)
28 tháng 4 2017
a)\(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2^2-1}+\dfrac{1}{4^2-1}+...+\dfrac{1}{100^2-1}\)
\(A< \dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)
\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}< \dfrac{50}{100}=\dfrac{1}{2}\)
Vậy \(A< \dfrac{1}{2}\)
b)B=\(\dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{2499}{2500}\)
49-B=\(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{2500}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)
\(49-B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(49-B< 1-\dfrac{1}{50}< 1\Leftrightarrow49< 1+B\Leftrightarrow B>48\)(ĐPCM)
28 tháng 4 2017
b) Đặt :
\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+............+\dfrac{2499}{2500}\)
\(\Rightarrow A=\dfrac{4}{4}-\dfrac{1}{4}+\dfrac{9}{9}-\dfrac{1}{9}+.........+\dfrac{2500}{2500}-\dfrac{1}{2500}\)
\(A=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+...........+1-\dfrac{1}{50^2}\)
\(A=\left(1+1+....+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)\)(\(49\) chữ số \(1\))
\(A=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+........+\dfrac{1}{50^2}\right)\)
Lại có :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)
Mà :
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}< 1\)
\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)>-1\)
\(\Rightarrow49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+............+\dfrac{1}{50^2}\right)>49-1\)\(=48\)
\(\Rightarrow A>48\) \(\rightarrowđpcm\)
Áp dụng tính chất : a^n - b^n chia hết cho a-b thì :
8^100-1 = (8^2)^50 - 1^50 chia hết cho 8^2-1 = 63
Mà 63 chia hết cho 9 => 8^100-1 chia hết cho 9
Tk mk nha