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\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{2^2-1}+\frac{1}{3^2-1}+...+\frac{1}{50^2-1}\)
\(=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{48}-\frac{1}{50}+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{50}-\frac{1}{51}\right)< \frac{1}{2}\left(1+\frac{1}{2}\right)=\frac{3}{4}\left(dpcm\right)\)
a) Ta thấy: 1/2^2<1/1.2
1/3^2<1/2.3
1/4^2<1/3.4
…………...
1/100^2<1/99.100
=>A<1/1.2+1/2.3+1/3.4+…+1/99.100=99/100
Mà 99/100<1 => 1/22 + 1/32 + 1/42 + ... + 1/1002<1
b)Ta thấy : 1/101+1/102+1/103+…+1/150>1/150+1/150+1/150+…+1/150(50 số hạng)
=>A>50/150>1/3 (1)
Ta thấy : 1/101+1/102+1/103+…+1/150<1/100+1/100+1/100+…+1/100(50 số hạng)
=>A<1/2 (2)
Từ (1) và (2) =>1/3<A<1/2
c) Ta thấy : 1/11 + 1/12 + 1/13 + ... + 1/20>1/20+1/20+1/20+…+1/20(10 số hạng)
=>1/11 + 1/12 + 1/13 + ... + 1/20>1/2
\(A=1+2+2^2+2^3+........+2^{2017}\)
\(2A=2+2^2+2^3+2^4+.......+2^{2018}\)
\(2A-A=\left(2+2^2+2^3+2^4+.......+2^{2018}\right)-\left(1+2+2^2+2^3+......+2^{2017}\right)\)
\(2A-A=2+2^2+2^3+2^4+........+2^{2018}-1-2-2^2-2^3-......-2^{2017}\)
\(\Rightarrow A=2^{2018}-1\)
Gọi tổng trên là A
Ta có :
\(A=1+2+2^2+2^3+...+2^{2017}\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2018}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2018}\right)-\left(1+2+2^2+2^3+...+2^{2017}\right)\)
\(\Leftrightarrow A=2^{2018}-1\)
Vậy \(A=2^{2018}-1\)
Thấy: với mọi n > 0 thì \(\frac{1}{2^n}<\frac{1}{\left(n-1\right)n}\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^n}\)
Ta cũng có:
\(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{2^3}<\frac{1}{2.3}\)
\(\frac{1}{2^4}<\frac{1}{3.4}\)
\(...\)
\(\frac{1}{2^n}<\frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^n}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}<1\)
\(\Rightarrow\)A<B<1
Vậy \(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^n}<1\)
Ta có : \(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{2^3}<\frac{1}{2.3}\)
\(\frac{1}{2^4}<\frac{1}{3.4}\)
...........
\(\frac{1}{2^n}<\frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^n}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}<1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^n}<1\)
ta có: \(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
\(\frac{1}{4^2}<\frac{1}{3.4}\)
...............
\(\frac{1}{n^2}<\frac{1}{\left(n-1\right).n}\)
cộng vế với vế ta được:
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(VP=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}=\frac{n}{n}-\frac{1}{n}=\frac{n-1}{n}<1\)
\(=>VP<1\)
\(\ \)Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}<1\left(dpcm\right)\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}< 1\left(đpcm\right)\)