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\(M=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}+\frac{1}{5^{2014}}\)
\(5M=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}\)
\(\Rightarrow4M=1-\frac{1}{5^{2014}}< 1\)
\(\Rightarrow M< \frac{1}{4}< \frac{1}{3}\)
Ta có :
M = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
3M = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
3M - M = ( \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)) - ( \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\))
2M = \(1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow M=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
3M=1+1/3+1/3^2+....+1/3^98
2M=3M-M=(1+1/3+1/3^2+....+1/3^98)-(1/3+1/3^2+....+1/3^99) = 1-1/3^99 < 1
=> M < 1/2
=> ĐPCM
k mk nha
Bài này dễ,ông không chịu làm thì có ^_^:
Ta có:\(B=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+....+\left(\frac{1}{2^{2014}+1}+....+\frac{1}{2^{2015}}\right)+\frac{1}{2^{2015}+1}+...+\frac{1}{2^{2016}-1}\)
\(>1+\frac{1}{2}+2.\frac{1}{2^2}+2^2.\frac{1}{2^3}+........+2^{2014}.\frac{1}{2^{2015}}\)
\(=1+\frac{1}{2}+\frac{1}{2}+.........+\frac{1}{2}\) (có 2015 phân số \(\frac{1}{2}\))
\(=1+2014.\frac{1}{2}+\frac{1}{2}=1008+\frac{1}{2}>1008\)
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{49}+\left(\frac{1}{5}\right)^{50}\)
\(5M=1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{48}+\left(\frac{1}{5}\right)^{49}\)
5M - M = \(1-\left(\frac{1}{5}\right)^{50}\)hay 4M = \(1-\left(\frac{1}{5}\right)^{50}\)< 1
\(\Rightarrow M=\frac{1-\left(\frac{1}{5}\right)^{50}}{4}< \frac{1}{4}\)
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{50}\)(1)
\(\Rightarrow5M=1+\frac{1}{5}+...+\left(\frac{1}{5}\right)^{49}\)(2)
Lấy (2)-(1) ta có
\(\Rightarrow4M=1-\left(\frac{1}{5}\right)^{50}\)
\(\Rightarrow M=\frac{1-\frac{1}{5^{50}}}{4}\)
Do \(1-\frac{1}{5^{50}}< 1\)
\(\Rightarrow M< \frac{1}{4}\)
sửa đề câu 1 :
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
sửa đề câu 2
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
\(A=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2010^2}>1-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}-\frac{1}{2010}=\frac{1004}{2010}>\frac{1}{2010}\Rightarrow A>\frac{1}{2010}\)
\(\frac{1}{M}=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{59.60}{2}}\)
\(\frac{1}{M}=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{59.60}\)
\(\frac{1}{M}=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{59}-\frac{1}{60}\right)\)
\(\frac{1}{M}=\frac{2}{3}-\frac{2}{60}< \frac{2}{3}\)
-theo t đề là M chứ ko phải 1/M