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\(\frac{1}{3^2}<\frac{1}{3.4}\)
\(\frac{1}{4^2}<\frac{1}{4.5}\)
\(\frac{1}{5^2}<\frac{1}{5.6}\)
\(...\)
\(\frac{1}{100^2}<\frac{1}{100.101}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{101}\)
Mà \(\frac{1}{3}<\frac{1}{2}\) nên \(\frac{1}{3}-\frac{1}{101}<\frac{1}{2}\)
hay \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{2}\)
Đặt A=1/3^2+1/4^2+1/5^2+...+1/100^2
Suy raA<1/2*3+1/3*4+1/4*5+..+1/99*100
A<1/2-1/100<1/2
Ta có điều phải chứng minh.
1/2^2>1/2.3;1/3^2>1/3.4;......;1/9^2>1/9.10
suy ra S > 1/2.3+1/3.4+......+1/9.10
S> 1/2-1/3+1/3-1/4 +.....+1/9-1/10
S> 1/2-1/10=2/5
Vay 2/5 < S
Vậy còn S < \(\frac{8}{9}\)thì sao, bạn quên chưa chứng minh rồi
1/2+1/4+1/8+1/16+1/32+1/64=32/64+16/64+8/64+4/64+2/32+1/64=63/64<1
Đặt \(B=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\)
Ta có : \(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
...
\(\frac{1}{2014^2}< \frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{2014}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2}=\frac{3}{4}\)
Vậy A<\(\frac{3}{4}\)
A<\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)=\(\frac{2013}{2014}\)<\(\frac{3}{4}\)
Ta có: A = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}<\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Rightarrow\) A < \(1+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow\) A < \(1+\left(1-\frac{1}{50}\right)\)
\(\Rightarrow\) A < 1 + 49/50
Mà 1+49/50 < 2 nên A < 1+49/50 < 2
\(\Rightarrow\) A < 2
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{21.23}\)
\(=5-\frac{5}{3}+\frac{5}{3}-\frac{5}{5}+\frac{5}{5}-\frac{5}{7}+...+\frac{5}{21}-\frac{5}{23}\)
\(=5\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)
\(=5\left(1-\frac{1}{23}\right)\)
\(=5.\frac{22}{23}\)
\(=\frac{110}{23}\)
\(A=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{21}-\frac{1}{23}\right)\)
\(A=\frac{5}{2}.\left(1-\frac{1}{23}\right)\)
\(A=\frac{5}{2}.\frac{22}{23}\)
\(A=\frac{55}{23}\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=1-1\frac{1991}{1993}=\frac{1991}{1993}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{1991}{1993}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{1993}:2=\frac{1991}{3986}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}=\frac{1}{1993}\)
=> x + 1 = 1993
=> x = 1993 - 1
=> x = 1992
\(A=\frac{\left(23\frac{11}{15}-26\frac{13}{20}\right)}{12^2+5^2}\cdot\frac{1-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+\frac{11}{15}-26+\frac{13}{20}\right)}{144+25}\cdot\frac{1-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}}{9.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+26+\frac{11}{15}-\frac{13}{20}\right)}{169}\cdot\frac{1-\left(\frac{1}{5}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)}{13.\left(9.2-5\right)}-\frac{19}{37}\)
\(A=\frac{49+\frac{44}{60}-\frac{39}{60}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}}{13.13}-\frac{19}{37}\)
\(A=\frac{49+\frac{1}{20}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{8}}{169}-\frac{19}{37}\)
\(A=\frac{49\frac{1}{20}}{169}\cdot\frac{\frac{4}{5}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{32}{40}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{37}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981.\frac{37}{40}}{169^2}-\frac{19}{37}\)
\(A=\frac{\frac{36297}{40}}{28561}-\frac{19}{37}\)
\(A=\frac{907,425}{28561}-\frac{19}{37}\)
\(A=\frac{33574,725}{1056757}-\frac{542659}{1056757}\)
\(A=\frac{-509084,275}{1056757}=-0,04604282...\)
Mik chỉ làm đc thế này thôi, ôn thi học kì II tốt nha bạn!
\(S=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{20^2}\)
\(S=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{20^2}<\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(S<\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(S<\frac{1}{2}-\frac{1}{20}<\frac{1}{2}\)
Vậy \(S<\frac{1}{2}\)
Cám ơn bạn rất nhiều hjhj