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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 )
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 \(A< \frac{1}{4}\)
\(A=\left(1+\frac{1}{3}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\)
\(A=\frac{3+1}{3}.\frac{8+1}{8}.\frac{15+1}{15}...\frac{n^2+2n+1}{n^2+2n}\)
\(A=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{\left(n+1\right)^2}{n^2+2n}\)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(A=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)
\(A=\left(n+1\right).\frac{2}{n+2}=\frac{2.\left(n+1\right)}{n+2}\)
Ta có : \(1+\frac{1}{k^2+2k}=\frac{k^2+2k+1}{k^2+2k}=\frac{\left(k+1\right)^2}{k\left(k+2\right)}\) với k thuộc N*
Áp dụng với k = 1,2,3,....,n được :
\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\)
\(=\frac{\left(1+1\right)^2}{1.\left(1+2\right)}.\frac{\left(2+1\right)^2}{2.\left(2+2\right)}.\frac{\left(3+1\right)^2}{3.\left(3+2\right)}...\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(=\frac{\left[2.3.4...\left(n+1\right)\right]^2}{1.2.3...n.3.4.5...\left(n+2\right)}=\frac{\left[\left(n+1\right)!\right]^2}{n!.\frac{\left(n+2\right)!}{2}}\)
Bạn tham khảo cách làm ở đây: https://olm.vn/hoi-dap/question/528628.html
Ta có :
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(A=\frac{4-1}{1^2.2^2}+\frac{9-4}{2^2.3^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}\)
\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}\)
\(A=\frac{2^2}{1^2.2^2}-\frac{1^2}{1^2.2^2}+\frac{3^2}{2^2.3^2}-\frac{2^2}{2^2.3^2}+...+\frac{\left(n+1\right)^2}{n^2\left(n+1\right)^2}-\frac{n^2}{n^2\left(n+1\right)^2}\)
\(A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
\(A=1-\frac{1}{\left(n+1\right)^2}\)
Chúc bạn học tốt ~
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(3.2\right)^2}+...+\frac{2n+1}{\left[n.\left(n+1\right)\right]^2}\)
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{2n+1}{n^2.\left(n+1\right)^2}\)
\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2.\left(n+1\right)^2}\)
\(A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\)
\(A=1-\frac{1}{\left(n+1\right)^2}\)
mk chỉ làm được đến đấy thôi
a) \(\frac{-32}{\left(-2\right)^n}=4\)
\(\frac{\left(-2\right)^5}{\left(-2\right)^n}=4\)
\(\left(-2\right)^{5-n}=\left(-2\right)^2\)
=> 5-n = 2
n = 3
b) \(\frac{8}{2^n}=2\)
\(\frac{2^3}{2^n}=2\)
\(2^{3-n}=2^1\)
=> 3 -n = 1
n = 2
c) \(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\)
\(\left(\frac{1}{2}\right)^{2n-1}=\left(\frac{1}{2}\right)^3\)
=> 2n -1 = 3
2n = 4
n = 2
a) \(\frac{-32}{\left(-2\right)^n}=4\Leftrightarrow\left(-2\right)^n=\frac{-32}{4}\)
\(\left(-2\right)^n=-8\)Mà \(-8=2^{-3}\)
\(\Rightarrow x=-3\)
b) \(\frac{8}{2^n}=2\Leftrightarrow2^n=\frac{8}{2}\)
\(2^n=4\) Mà \(4=2^2\Rightarrow x=2\)
c) \(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\Rightarrow\left(\frac{1}{2}\right)^{2n}:\frac{1}{2}=\frac{1}{8}\)
\(\left(\frac{1}{2}\right)^{2n}=\frac{1}{8}\cdot\frac{1}{2}\)
\(\left(\frac{1}{2}\right)^{2n}=\frac{1}{16}\Leftrightarrow\frac{1}{2^{2n}}=\frac{1}{16}\) mà\(16=2^4\)
\(2n=4\Rightarrow n=2\)
Vậy .........................