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Đặt A là tên biểu thức
\(A=1-\frac{15}{16}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{4n^2}\)
\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2^2n^2}\)
\(A=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
\(A< \frac{1}{2^2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\right)\)
\(A< \frac{1}{2^2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(A< \frac{1}{2^2}\left(1-\frac{1}{n}\right)=\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)(đpcm)
Ta có: 9A=1+1/32+...+1/398
Vậy 10A=(1+1/32+...+1/398) + (1/32+1/34+...+1/3100)
10A=1+2(1/32+1/34+...+1/398)+1/3100
Vậy 10A>1 suy ra A > 0,1 suy ra người ra đề đã đặt sai đề!
\(A=\frac{1}{3^2}-\frac{1}{3^4}+....+\frac{1}{3^{4n-2}}-\frac{1}{3^{4n}}+....+\frac{1}{3^{98}}+\frac{1}{3^{100}}\)
Suy ra \(3^2.A=1-\frac{1}{3^2}+.....+\frac{1}{3^{4n-4}}-\frac{1}{3^{4n-2}}+...+\frac{1}{3^{96}}-\frac{1}{3^{98}}\)
Khi đó \(3^2.A-A=1-\frac{1}{3^{100}}\)hay \(8A=1-\frac{1}{3^{100}}\Rightarrow A=\frac{1}{8}-\frac{1}{3^{100}}< 0,1\)
Vậy
Ta có :
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{4n^2}=\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+\frac{1}{\left(2.4\right)^2}+...+\frac{1}{\left(2.2n\right)^2}\)
\(=\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+\frac{1}{2^2.4^2}+...+\frac{1}{2^2.4n^2}=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{4n^2}\right)\)
\(< \frac{1}{2^2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(4n^2-1\right)4n^2}\right)\)
\(=\frac{1}{4}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{4n^2-1}-\frac{1}{4n^2}\right)=\frac{1}{4}\left(1-\frac{1}{4n^2}\right)\)
\(=\frac{1}{4}-\frac{1}{16n^2}< \frac{1}{4}\) ( vì \(\frac{1}{16n^2}>0\) )
Vậy \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{4n^2}< \frac{1}{4}\)
Chúc bạn học tốt ~
=> 22.S = \(1-\frac{1}{2^2}+\frac{1}{2^4}-............+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}\)
=> 4S + S = \(1-\frac{1}{2^2}+\frac{1}{2^4}-......+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}+\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-....+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)
=> 5S = \(1-\frac{1}{2^{2004}}<1\)
=> S < 1 : 5 = 0,2 (đpcm)