Bài 1:
chug to : N = \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+.....+\frac{1}{\left(2n^2\right)}\) <\(\frac{1}{4}\)
P = \(\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+.....+\frac{2!}{n!}\)< 1
\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...........+\frac{1}{\left(2n\right)^2}< 4\left(v\text{ới}n\in N;n\ge2\right)\)
Đề là chứng minh N < 1/4 sẽ đúng hơn
Ta có :
\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(\Rightarrow2^2.N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
Ta lại có :
\(4N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}\)
\(\Rightarrow N< \left(1-\frac{1}{n}\right):4=\frac{1}{4}\left(1-\frac{1}{n}\right)\)
Mà \(n\in N;n\ge2\)=> 1 -\(\frac{1}{n}\)< 1
=> \(N< \frac{1}{4}\left(1-\frac{1}{n}\right)< \frac{1}{4}\)
=> \(N< \frac{1}{4}\)( đpcm )
Thank you very much
Cho \(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\left(n\in N,n.2\right)\)
Chứng minh A<1/4
Ta có :
\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(A=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\right)\)
\(A< \frac{1}{4}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{4}\left(1-\frac{1}{n}\right)\)
\(A< \frac{1}{4}-\frac{1}{4n}\)
Lại có \(n>0\) nên \(\frac{1}{4n}>0\)
\(\Rightarrow\)\(\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)
Vậy \(A< \frac{1}{4}\)
\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+.......+\frac{1}{\left(2n\right)^2}< 4\left(v\text{ới}n\in N;n\ge2\right)\)\(2\)
CMR \(\forall n\in\)N* ta có
\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+...+\left(\frac{1}{2n-1}-\frac{1}{2n}\right)=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\)
CMR: \(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)(n thuộc N , n lớn hơn bằng 2)
Ta có :
\(N=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
Ta thấy : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
.......
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1\)
\(\Rightarrow\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)< 1.\frac{1}{2^2}\)
\(\Rightarrow N< \frac{1}{4}\)(ĐPCM)
Ủng hộ mk nha !!! ^_^
1. Chứng minh : B = \(\left(1-\frac{2}{6}\right).\left(1-\frac{2}{12}\right).\left(1-\frac{2}{20}\right)...\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\)
2. cho M = \(\frac{1}{1.\left(2n-1\right)}+\frac{1}{3.\left(2n-3\right)}+\frac{1}{5.\left(2n-5\right)}+...+\frac{1}{\left(2n-3\right).3}+\frac{1}{\left(2n-1\right).1}\)
N = \(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-1}\)
Rút gọn \(\frac{M}{N}\)
CMR với mọi số tự nhiên n>2 thì :
a)\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)<\(\frac{1}{2}\)
b)\(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}\)<\(\frac{1}{4}\)
c)\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{\left(2n+1\right)^2}\right)\)<2
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
1) CMR :
M=\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)với n thuộc N và n\(\ge\)2
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{\left(2n-2\right)2n}\)
\(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}< \frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{\left(2n-2\right)2n}\)\(.\frac{1}{2}\) Ta gọi là A
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{\left(2n-2\right)2n}\right)\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2n}\right)=\frac{1}{2}.\frac{1}{2}-\frac{1}{2}.\frac{1}{2n}=\frac{1}{4}-\frac{1}{2n.2}\)
\(\Rightarrow M< \frac{1}{4}-\frac{1}{2n.2}< \frac{1}{4}\)
\(\Rightarrow M< \frac{1}{4}\left(Đpcm\right)\)
\(\)