Chứng tỏ rằng : \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+.......+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
nhanh lên nha các bạn iu
Chứng tỏ : \(C=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
Ta có:
\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)
\(\frac{1}{6^2}=\frac{1}{6.6}< \frac{1}{5.6}\)
\(\frac{1}{8^2}=\frac{1}{8.8}< \frac{1}{7.8}\)
\(...\)
\(\frac{1}{\left(2n\right)^2}=\frac{1}{2n.2n}< \frac{1}{1n.2n}\)
Vậy \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)\(< \)\(\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{1n.2n}\)
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)\(< \)\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{1n}-\frac{1}{2n}\)
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)\(< \)\(\frac{1}{3}+\left(\frac{-1}{4}+\frac{1}{4}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+...-\frac{1}{2n}\)
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)\(< \)\(\frac{1}{3}-\frac{1}{2n}\)
1. Chứng minh :
A = \(\frac{^{^{ }}1}{4^2}+\frac{1}{^{ }6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n^2\right)}< \frac{1}{4}\) ( n E N ; n < 2 )
Nhanh LÊN NHA MK CẦN GẤP GẤP LẮM LẮM LUÔN LUÔN ĐÓ
Ai nhanh mk cho 1 TTIICCK
\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2n-2\right).2n}\)
\(< \frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2n-2}-\frac{1}{2n}\right)\)
\(< \frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n}\right)=\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)
\(\Rightarrow\) \(A< \frac{1}{4}\)
Study well ! >_<
Chứng minh rằng: \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}
CHỨNG MINH RẰNG:
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
Đặt \(A=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
Có:
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(...\)
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}< 1\)
\(\Rightarrow A< \frac{1}{2^2}.1=\frac{1}{4}\)
CHỨNG MINH RẰNG:
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+...+\frac{1}{\left(2n^2\right)}< \frac{1}{4}\)
=>\(\frac{1}{2^2}\)x (\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{n^2}\))
Đặt A=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{n^2}\)
Ta có:\(\frac{1}{2^2}\)<\(\frac{1}{1\cdot2}\)
\(\frac{1}{3^2}\)<\(\frac{1}{2\cdot3}\)
.........\(\frac{1}{n^2}\)<\(\frac{1}{\left(n-1\right)\cdot n}\)
=>\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{n^2}\)<\(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+...+\(\frac{1}{\left(n-1\right)\cdot n}\)
=>A<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{n-1}\)--\(\frac{1}{n}\)
=>\(\frac{1}{2^2}\)*A<\(\frac{1}{2^2}\)(1--\(\frac{1}{n}\))
=>\(\frac{1}{2^2}\)*A<\(\frac{1}{4}\)(1--\(\frac{1}{n}\))
=>\(\frac{1}{2^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{\left(2n\right)^2}\)<\(\frac{1}{4}\)--\(\frac{1}{4n}\)<\(\frac{1}{4}\)
=>\(\frac{1}{2^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{\left(2n\right)^2}\)<\(\frac{1}{4}\)
Chứng minh rằng: \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}
đặt A=1/2^2+1/4^2+1/6^2+.....+1/(2n)^2
ta có :
A=1/2^2 +1/2^2(1/2^2+1/3^2+1/4^2+.....+1/n^2)
A<1/2^2+1/2^2(1/1.2+1/2.3+...+1/(n-1)n)
=1/2^2+1/2^2(1-1/2+1/2-1/3+....+1/(n-1)-1/n)
=1/2^2+1/2^2(1-1/n)
<1/2^2+1/2^2.1=1/2<3/4
vậy A<3/4
Chứng minh rằng: \(M=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}<\frac{1}{4}\)
Ta có:
\(M=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(=\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+\frac{1}{2^2.4^2}+...+\frac{1}{2^2.n^2}\)
\(=\frac{1}{2^2}.\frac{1}{2^2}+\frac{1}{2^2}.\frac{1}{3^2}+\frac{1}{2^2}.\frac{1}{4^2}+...+\frac{1}{2^2}.\frac{1}{n^2}\)
\(=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)
Coi \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{n.n}\)
Vì: \(\frac{1}{2.2}
What? lớp 5 mà học cả số mũ?? thời nay bọn trẻ con học trâu thật!
Cho $A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\left(n\in Z;n\ge2\right)$A=142 +162 +182 +...+1(2n)2 (n∈Z;n≥2)
Chứng tỏ A$\notin$∉ N
Chứng minh :\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\).... \(+\frac{1}{\left(2n\right)^2}\)= \(\frac{1}{2^2}\). ( \(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{n^2}\)) < \(\frac{1}{2^2}\)( \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).\left(n\right)}\)) = \(\frac{1}{2^2}\)( \(1-\frac{1}{n}\)) < \(\frac{1}{2^2}\).1 = \(\frac{1}{4}\)
\(\Rightarrow\)\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)< \(\frac{1}{4}\)