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Ta có : \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\)
Mà \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{8^2}<\frac{1}{7.8}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=1-\frac{1}{8}<1\)
Vậy B < 1
Bạn xem lời giải của mình nhé:
Giải:
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{6.7}+\frac{1}{7.8}\\\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8} \\ =\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}< 1\\ \Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< 1\)
Chúc bạn học tốt!
Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{7.8}\)
= \(1-\frac{1}{8}< 1\)
Vậy B < 1
\(S=7(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{61}-\frac{1}{63}) \)
\(S=7(\frac{1}{3}-\frac{1}{63})\)
\(S=7(\frac{21}{63}-\frac{1}{63}) \)
\(S=7.\frac{20}{63}\)
\(S=\frac{20}{9}\)
Do đó:\(S<\frac{5}{2}\)
S=\(\frac{2.7}{3.5}+\frac{2.7}{5.7}+\frac{2.7}{7.9}+....+\frac{2.7}{61.63}\)và\(\frac{5}{2}\)
S=7.(\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....-\frac{1}{63}\)) và\(\frac{5}{2}\)
S=7.(\(\frac{1}{3}-\frac{1}{63}\)) và\(\frac{5}{2}\)
S=7.\(\frac{20}{63}\)và\(\frac{5}{2}\)
=>S=\(\frac{20}{9}\)so với \(\frac{5}{2}\)
=>S=\(\frac{40}{18}\)và\(\frac{45}{18}\)
=>S<\(\frac{5}{2}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{5}{6}\) -\(\frac{3}{4}\) + \(\frac{2}{3}\) -\(\frac{1}{2}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{10}{12}\)-\(\frac{9}{12}\)+\(\frac{8}{12}\)-\(\frac{6}{12}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\)= \(\frac{1}{4}\)=> x. (\(\frac{1}{2}\)- \(\frac{2}{3}\) + \(\frac{3}{4}\)- \(\frac{5}{6}\)) = \(\frac{1}{4}\)=> x.( \(\frac{6}{12}\)- \(\frac{8}{12}\)+\(\frac{9}{12}\)-\(\frac{10}{12}\))= \(\frac{1}{4}\)=> x. \(\frac{-1}{4}\)=\(\frac{1}{4}\)=> x = \(\frac{1}{4}\): \(\frac{-1}{4}\)=> x = -1=>x.(1/2-2/3+3/4)=1/4
=>x.7/12=1/4
=>x=1/4:7/12
=>x=1/4.12/7
=>x=3/7
\(S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-....+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)
\(<\frac{1}{2^4}-\frac{1}{2^4}+\frac{1}{2^8}-\frac{1}{2^8}+...+\frac{1}{2^{4n}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2004}}-\frac{1}{2^{2004}}\)=0+0+0+...+0+....+0=0 <0,2
Vậy S<0,2
lồn
Ta có:
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(=\frac{1}{4}+\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)\)
Đặt \(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(B=\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}\right)+\left(\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)\)
Giả sử tất cả các số hạng của B đều bằng \(\frac{1}{6^2}\)
\(\Rightarrow B=6.\frac{1}{6^2}=\frac{6}{36}=\frac{1}{6}<\frac{1}{4}\)
Do đó \(B<\frac{1}{4}\)
\(\Rightarrow A=\frac{1}{4}+B<\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
Vậy \(A<\frac{1}{2}\)