vì: \(\dfrac{1}{4^2}< \dfrac{1}{4}\)
\(\dfrac{1}{6^2}< \dfrac{1}{4}\)
........
\(\dfrac{1}{2020^2}< \dfrac{1}{4}\)
=> \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+.......+\dfrac{1}{2020^2}< \dfrac{1}{4}\)
\(A=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{2020}}-\dfrac{1}{2^{2022}}
\)
Chứng minh A<0.2
chứng minh rằng \(S=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(n\in N,n\ge2\right)\)
Bài 1:Chứng tỏ rằng:B=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)\(\dfrac{1}{8^2}\)<1
Bài 2:Chứng tỏ rằng:E=\(\dfrac{3}{4}\)+\(\dfrac{8}{9}\)+\(\dfrac{15}{16}\)+...+\(\dfrac{2499}{2500}\)<1
Bài 3:Chứng tỏ rằng:1<\(\dfrac{2011}{2020^2+1}\)+\(\dfrac{2021}{2020^2+2}\)+\(\dfrac{2021}{2020^3+3}\)+...+\(\dfrac{2021}{2020^3+2020}\)< 2
cho A=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+.....+\(\dfrac{1}{2^{2020}}\)+\(\dfrac{1}{2^{2021}}\). Chứng tỏ rằng A<\(\dfrac{1}{2}\)
Giúp vs ạ cần gấp
*Thực hiện
1/ (\(\dfrac{2021}{2020}\)+\(\dfrac{2020}{2021}\)) x (\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\))
2/ (\(\dfrac{7}{19}\)-\(\dfrac{5}{12}\)):\(\dfrac{-5}{8}\)-(\(\dfrac{7}{19}\)-\(\dfrac{29}{12}\)):\(\dfrac{5}{8}\)
3/ \(\dfrac{-5}{6}\)x\(\dfrac{7}{24}\)-\(\dfrac{5}{6}\)x\(\dfrac{14}{24}\)-\(\dfrac{5}{6}\)x\(\dfrac{3}{24}\)
chứng minh rằng : \(\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
\(\dfrac{2}{5^2}+\dfrac{2}{6^2}+\dfrac{2}{7^2}+...+\dfrac{2}{2020^2}\)
Chứng minh dãy phân số trên < \(\dfrac{1}{2}\)
Chứng Minh Rằng :A=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{6^2}\)+...+\(\dfrac{1}{1002^2}\)<\(\dfrac{1}{2}\)
