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Đề bài sai rồi!Riêng 1/(2.2) đã bằng 1/4 rùi thì tổng trên phải lớn hơn 1/4 chứ!
Bạn Phạm Gia Bảo nói đúng đấy
Bạn nên sửa đề bài đi
\(G< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{199.200}\)
\(G< \frac{1-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{200-199}{199.200}\)
\(G< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(G< 1-\frac{1}{200}< 1\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)
\(2A+A=\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\right)\)
\(3A=1-\frac{1}{64}\)
\(3A=\frac{63}{64}\Rightarrow A=\frac{63}{64}\div3=\frac{21}{64}< \frac{1}{3}\)
Bài làm
Ta đặt M=1/3+1/7+1/13+1/21+1/31+1/43+1/57+1/73+1/91
Vậy M<1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90
M< 1/2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10
M< (1-1/2) +(1/2-1/3) +(1/3-1/4) +(1/4-1/5) +(1/5-1/6) +(1/6-1/7) +(1/7-1/8) +(1/8-1/9) +(1/9-1/10)
M< 1-1/10 < 9/10 (1)
Vì 9/10 < 1 (2)
Từ(1) và (2) ta có : 1/3+1/7+1/13+1/21+1/31+1/43+1/57+1/73+1/91<1
Thấy 1/41+1/42 +......+ 1/60 < 1/40 .20
1/41 +1/42 + .....+1/60<1/2
mà 1/61 +1/62+......+1/80 < 1/60 .20 =1/3
suy ra 1/41+1/42+ .......+1/80 <1/2 +1/3=7/12(đpcm)
Lại có 1/41 +1/42 +.....+1/80 <1/40 .40 =1(đpcm)
Ta có :
\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\Rightarrow A< \frac{8}{9}\)(1)
Lại có \(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\Rightarrow A>\frac{2}{5}\)(2)
Từ (1) (2) => \(\frac{2}{5}< A< \frac{8}{9}\left(\text{ĐPCM}\right)\)
Bài làm :
Ta có :
\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A>\frac{1}{2}-\frac{1}{10}\)
\(A>\frac{2}{5}\left(1\right)\)
Ta cũng có :
\( A=\frac{1}{2.2}+\frac{1}{3.3}+......+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{8.9}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-......+\frac{1}{8}-\frac{1}{9}\)
\(A< 1-\frac{1}{9}\)
\(A< \frac{8}{9}\left(2\right)\)
\(\text{Từ (1) và (2) }\Rightarrow\frac{2}{5}< A< \frac{8}{9}\)
=> Điều phải chứng minh
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Bài làm:
Ta có: \(S=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}\)
\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)\(\Rightarrow\frac{2}{5}< S\)
Cái còn lại tự CM
A= 1/2.2 + 1/3.3 + 1/4.4 + 1/5.5 + 1/6.6 + 1/7.7 + 1/8.8 + 1/9.9
Vì 1/2.2 > 1/2.3; 1/3.3 > 1/3.4 ; 1/5.5 > 1/5.6;...... nên
1/2.2 +1/3.3 + 1/4.4 + 1/5.5 + 1/6.6 + 1/7.7 + 1/8.8 + 1/9.9 > 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10
Ta có: 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10
= 1/2-1/3 + 1/3 -1/4 + 1/4-1/5+...+1/9-1/10
= 1/2- 1/10
= 2/5
Vì A < 2/5 mà 2/5 <7/8 nên 2/5 < A < 7/8
Vậy....
Giúp mình bài này với
\(\frac{1}{2.2}+\frac{1}{4.4}+\frac{1}{6.6}+...+\frac{1}{200.200}\)
\(=\frac{1}{4}\left(1+\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)\)
\(< \frac{1}{4}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=\frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{4}\left(1+1-\frac{1}{100}\right)=\frac{1}{4}\left(2-\frac{1}{100}\right)=\frac{1}{2}-\frac{1}{400}< \frac{1}{2}\)