Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{n^3}< \frac{1}{\left(n-2\right)n\left(n+1\right)}\Leftrightarrow\frac{\left(n-2\right)n\left(n+1\right)}{n^3}< 1\Leftrightarrow\left(n-2\right)\left(n+1\right)< n^2\)
\(\Leftrightarrow n^2-n-2< n^2\Leftrightarrow-n-2< 0\)Đúng \(\forall n\inℕ\)
--->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 \(A< \frac{1}{4}\)
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{50}\)
\(M=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{50}}\)
\(5M=5\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{50}}\right)\)
\(5M=1+\frac{1}{5}+...+\frac{1}{5^{49}}\)
\(5M-M=\left(1+\frac{1}{5}+...+\frac{1}{5^{49}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{50}}\right)\)
\(4M=1-\frac{1}{5^{50}}\)
\(M=\frac{1-\frac{1}{5^{50}}}{4}< \frac{1}{4}=0,25\)
Đpcm
1. D= 1/3 + 1/3.4 + 1/3.4.5 + 1/3.4.5....n < 1/2 + 1/3.4 + 1/4.5 + ...+ 1/ n.(n-1)
=> còn lại thì bạn có thể tự chứng minh
Ta có
\(A=\frac{1}{14}+\frac{1}{29}+...+\frac{1}{n^2+\left(n+1\right)^2+\left(n+2\right)^2}+...+\frac{1}{1877}\)
\(=\frac{1}{1^2+2^2+3^2}+\frac{1}{2^2+3^2+4^2}+...+\frac{1}{n^2+\left(n+1\right)^2+\left(n+2\right)^2}+...+\frac{1}{24^2+25^2+26^2}\)
\(B=n^2+\left(n+1\right)^2+\left(n+2\right)^2=3n^2+6n+5\left(1\right)\)
+ Với \(n\ge1\)từ (1) ta có \(B\le3n^2+9n+6=3\left(n^2+3n+2\right)=3\left(n+1\right)\left(n+2\right)\)Từ đó
\(A>\frac{1}{3}\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\right)=\frac{1}{3}C\)
Với \(C=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{25}-\frac{1}{26}=\frac{1}{2}-\frac{1}{26}=\frac{6}{13}\)
\(\Rightarrow A>\frac{1}{3}\cdot\frac{6}{13}=\frac{2}{13}>0,15\)
+ Với \(n\ge1\)từ (1) ta có \(B>2n^2+6n+4=2\left(n^2+3n+2\right)=2\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A< \frac{1}{2}\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\right)=\frac{1}{2}C\)
Với \(C=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{25}-\frac{1}{26}=\frac{1}{2}-\frac{1}{26}=\frac{6}{13}\)
\(\Rightarrow A< \frac{1}{2}\cdot\frac{6}{13}=\frac{3}{13}< 0,25\)
Vậy \(0,15< A< 0,25\)