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\(S=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{20^2}\)
\(S=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{20^2}<\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(S<\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(S<\frac{1}{2}-\frac{1}{20}<\frac{1}{2}\)
Vậy \(S<\frac{1}{2}\)
M=1/10 + 1/15 + 1/21 +....+ 1/120
M=2/20 +2/30+2/42+....+2/240
M=2/4.5 + 2/5.6 + 2/6.7 +.....+ 2/15.16
M=2.(1/4.5 +......+ 1/15.16)
M=2.(1/4 -1/5 +1/5 - 1/6 +.....+ 1/15 - 1/16)
M=2.(1/4 - 1/16)
M=2.(4/16 - 1/16)
M=2. 3/16
M=6/16=3/8
Có 1/3 = 8/24 < 9/24 = 3/8 =>1/3<M
Có 1/2 = 4/8>3/8 =>1/2 >M
=> 1/3 < M < 1/2
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<\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)=\(\frac{2013}{2014}\)<\(\frac{3}{4}\)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}\)
\(S>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(S>\frac{1}{2}-\frac{1}{10}\)
\(S>\frac{4}{10}=\frac{2}{5}\)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9.10}< S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3\cdot4}+...+\frac{1}{8.9}\)
=>\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}< S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{8}-\frac{1}{9}\)
=> \(\frac{1}{2}-\frac{1}{10}< S< 1-\frac{1}{9}\)
=> \(\frac{2}{5}< S< \frac{8}{9}\)(dpcm )
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}< 1\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}< 1\)
\(S=1-\frac{1}{50}< 1\)
\(S=\frac{49}{50}< 1\left(đpcm\right)\)
Ta có : S =\(\frac{1}{2^2}\)\(+\)\(\frac{1}{3^2}\)\(+\)...\(+\)\(\frac{1}{9^2}\)
= \(\frac{1}{2.2}\)\(+\)\(\frac{1}{3.3}\)\(+\)...\(+\)\(\frac{1}{9^2}\)
\(\Rightarrow\)S > \(\frac{1}{2.3}\)\(+\)\(\frac{1}{3.4}\)\(+\)...\(+\)\(\frac{1}{9.10}\)
= \(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)..\(+\)\(\frac{1}{9}\)\(-\)\(\frac{1}{10}\)
= \(\frac{1}{2}\)\(-\)\(\frac{1}{10}\)
\(\Rightarrow\)S < \(\frac{1}{1.2}\)\(+\)\(\frac{1}{2.3}\)\(+\)...\(+\)\(\frac{1}{8.9}\)
=\(1\)\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)...\(+\)\(\frac{1}{8}\)\(-\)\(\frac{1}{9}\)
= \(1\)\(-\)\(\frac{1}{9}\)= \(\frac{8}{9}\)
Vậy \(\frac{2}{5}\)< S < \(\frac{8}{9}\)(đpcm)
Chúc bạn học tốt
\(\frac{1}{3^2}<\frac{1}{3.4}\)
\(\frac{1}{4^2}<\frac{1}{4.5}\)
\(\frac{1}{5^2}<\frac{1}{5.6}\)
\(...\)
\(\frac{1}{100^2}<\frac{1}{100.101}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{101}\)
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.
Sorry bạn nha , mình bấm nhầm nút
\(A=\frac{5}{4}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A< \frac{5}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A< \frac{5}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< \frac{5}{4}+\frac{1}{2}-\frac{1}{100}< \frac{5}{4}+\frac{1}{2}=\frac{7}{4}\)
\(\Rightarrow\)\(A< \frac{7}{4}\)
Vậy , \(\frac{5}{4}< A< \frac{7}{4}\left(ĐPCM\right)\)
BÀI KHÓ CỦA TRƯỜNG MÌNH ĐÓ THI HK2
GIÚP MÌNH NHÉ!!!!!!THANKS!!!!!!
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