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\(M=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}+\frac{1}{5^{2014}}\)
\(5M=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}\)
\(\Rightarrow4M=1-\frac{1}{5^{2014}}< 1\)
\(\Rightarrow M< \frac{1}{4}< \frac{1}{3}\)
\(M=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{10}{3^{11}}\)
\(\Rightarrow3M=\frac{1}{3}+\frac{2}{3^2}+...+\frac{10}{3^{10}}\)
\(\Rightarrow3M-M=\left(\frac{1}{3}+\frac{2}{3^2}+...+\frac{10}{3^{10}}\right)-\left(\frac{1}{3^2}+\frac{2}{3^3}+...+\frac{10}{3^{11}}\right)\)
\(\Rightarrow2M=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}-\frac{10}{3^{11}}\)
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}\)
\(\Rightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^9}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^9}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^{10}}< 1\)
\(\Rightarrow2A< 1\)
\(\Rightarrow A< \frac{1}{2}\)
\(\Rightarrow2M< \frac{1}{2}-\frac{10}{3^{11}}\)
\(\Rightarrow M< \frac{\frac{1}{2}-\frac{10}{3^{11}}}{2}\)
\(\Rightarrow M< \frac{1}{4}-\frac{1}{2.3^{11}}< \frac{1}{4}\)
\(\Rightarrow M< \frac{1}{4}\left(đpcm\right)\)
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{49}+\left(\frac{1}{5}\right)^{50}\)
\(5M=1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{48}+\left(\frac{1}{5}\right)^{49}\)
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
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
\(\frac{1}{M}=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{59.60}{2}}\)
\(\frac{1}{M}=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{59.60}\)
\(\frac{1}{M}=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{59}-\frac{1}{60}\right)\)
\(\frac{1}{M}=\frac{2}{3}-\frac{2}{60}< \frac{2}{3}\)
-theo t đề là M chứ ko phải 1/M