cho n lớn hơn hoặc bằng 2. So sánh: B=1/2^2+1/4^2+1/6^2+...+1/(2n)^2 vs 1/2
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a) \(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)
Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}< 1\)( vì n \(\ge\)2 )
\(\Rightarrow N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}.1=\frac{1}{4}\)
Vậy \(N< \frac{1}{4}\)
b) \(P=\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+...+\frac{2!}{n!}\)
\(P=2!\left(\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+...+\frac{1}{n!}\right)\)
\(P< 2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(n-1\right).n}\right)\)
\(P< 2.\left(\frac{1}{2}-\frac{1}{n}\right)=1-\frac{2}{n}< 1\)
Vậy \(P< 1\)
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Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
...
\(\frac{1}{n^2}< \frac{1}{n\left(n-1\right)}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n-1\right)}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(A< 1-\frac{1}{n}< 1\) (vì \(n\ge2\))
Vậy \(A< 1\).
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A=1/2^2(1/2^2+1/3^2+...+1/n^2)<1/4[(1/(1.2)+1/(2.3)+...+1/(n-1).n]=1/4(1-1/n) {n lon hon hoac bang 2}. Suy ra 1-1/n<0. Suy ra A<1/4
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A=1/4^2+1/6^2+...+1/(2n)^2
=1/4(1/2^2+1/3^2+...+1/n^2)
=>A<1/4(1-1/2+1/2-1/3+...+1/n-1-1/n)
=>A<1/4(1-1/n)<1/4
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nhanh lên nhé các bạn trả lời nhanh và đúng thì mình tích cho
\(B=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(2B=\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}+\frac{1}{\left(2n\right)^2}\)
\(< \frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n-1\right)^2}+\frac{1}{\left(2n\right)^2}\)
\(< \frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{\left(2n-2\right)\left(2n-1\right)}+\frac{1}{\left(2n-1\right)2n}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2n-1}-\frac{1}{2n}\)
\(=1-\frac{1}{2n}< 1\)
Suy ra \(B< \frac{1}{2}\).