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\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)
\(2A=1-\frac{1}{3^8}\)
\(A=\frac{1}{2}-\frac{1}{2.3^8}\)
#)Giải : (Đg rảnh nên làm lun :v)
Ta có : \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}=1-\frac{1}{51}=\frac{50}{51}< 2\)
\(\Rightarrow A< \frac{50}{51}< 2\)
\(\Rightarrow A< 2\left(đpcm\right)\)
#)Giải :
\(\left(2011.2012+2012.2013\right).\left(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\right)\)
\(=\left(2011.2012+2012.2013\right).\left(1+\frac{1}{3}-1\frac{1}{3}\right)\)
\(=\left(2011.2012+2012.2013\right).\left(1\frac{1}{3}-1\frac{1}{3}\right)\)
\(=\left(2011.2012+2012.2013\right).0\)
\(=0\)
#~Will~be~Pens~#
\(P=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{999}\right)\left(1-\frac{1}{1000}\right)\)
\(P=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{998}{999}\cdot\frac{999}{1000}\)
\(P=\frac{1\cdot2\cdot3\cdot4\cdot...\cdot999}{2\cdot3\cdot4\cdot5\cdot...\cdot1000}\)
\(P=\frac{1}{1000}\)
\(P=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{998}{999}\times\frac{999}{1000}\)
P=1/1000
_Kudo_