\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+..+\frac{1}{9^2}\)
CMR \(\frac{2}{5}< S< \frac{8}{9}\)
Cho \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{9^2}\)
CMR : 2/5 < S < 8/9
CHO S=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+..........+\(\frac{1}{9^2}\)
CMR;\(\frac{2}{5}\)<S<\(\frac{8}{9}\)
\(S=\frac{1}{2}:\frac{3}{2}:\frac{4}{3}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}:\frac{9}{8}:\frac{10}{9}\)
\(S=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.\frac{9}{10}\)
\(S=\frac{1}{10}\)
học tốt
S=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{9}{10}\)
S=\(\frac{1.2.3....9}{2.3.4...10}=\frac{1}{10}\)
Vậy S=\(\frac{1}{10}\)
\(S=\frac{1}{2}:\frac{3}{2}:\frac{4}{3}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}:\frac{9}{8}:\frac{10}{9}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.\frac{9}{10}\)
\(=\frac{1.2.3.4.5.6.7.8.9}{2.3.4.5.6.7.8.9.10}\)
\(=\frac{1}{10}\)
Cho S=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{9^2}\)
Chứng minh rằng: \(\frac{2}{5}< S< \frac{8}{9}\)
Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}.\)
Mà\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)
\(\Rightarrow S< \frac{8}{9}\)
Và \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}\)
Mà \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow S>\frac{2}{5}\)
Vậy: \(\frac{2}{5}< S< \frac{8}{9}\)
Cho \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{9^2}\)
Chứng minh \(\frac{2}{5}< 5< \frac{8}{9}\)
Giải rõ ra
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
=> \(S=< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
=> \(S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)
=> \(\frac{2}{5}< S< \frac{8}{9}\)
Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=\frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{9}=\frac{9}{36}+\frac{18}{36}-\frac{4}{36}=\frac{9+18+\left(-4\right)}{36}=\frac{23}{36}< \frac{32}{36}=\frac{8}{9}\left(1\right)\)
Lại có: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2^2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{4}+\frac{1}{3}-\frac{1}{10}=\frac{15}{60}+\frac{20}{60}-\frac{6}{60}=\frac{19}{60}>\frac{8}{20}\left(2\right)\)
Từ (1) và (2) => đpcm
Cho \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
Chứng minh rằng \(\frac{2}{5}\)\(< S< \frac{8}{9}\)
Nhanh + đúng = tick nhé
( 99 - 1 ) : 2 + 1 = 50 ( số )
làm bừa thui,ai tích mình mình tích lại
Số số hạng là :
Có số cặp là :
50 : 2 = 25 ( cặp )
Mỗi cặp có giá trị là :
99 - 97 = 2
Tổng dãy trên là :
25 x 2 = 50
Đáp số : 50
Ta có : \(S>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(\Rightarrow S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow S>\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\) (1)
Ta lại có : \(S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)
\(\Rightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(\Rightarrow S< 1-\frac{1}{9}=\frac{8}{9}\)
Từ (1) và (2) \(\Rightarrow\frac{2}{5}< S< \frac{8}{9}\) ( đpcm )
Ta có :
22 = 2.2 < 2.3 => \(\frac{1}{2^2}>\frac{1}{2.3}\)
32 = 3.3 < 3.4 => \(\frac{1}{3^2}>\frac{1}{3.4}\)
........................
92 = 9 . 9 < 9. 10 => \(\frac{1}{9^2}>\frac{1}{9.10}\)
=> S > \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
=> 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 >\(\frac{2}{5}\)( 1 )
Ta có :
22 = 2 . 2 > 1.2 => \(\frac{1}{2^2}< \frac{1}{1.2}\)
32 = 3.3 > 3.2 => \(\frac{1}{3^2}< \frac{1}{2.3}\)
...........................
92 = 9.9 > 8.9 => \(\frac{1}{9^2}< \frac{1}{8.9}\)
=> S < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}\)
=> S < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{8}-\frac{1}{9}\)
=> S < \(1-\frac{1}{9}\)
=> S <\(\frac{8}{9}\)( 2 )
Từ ( 1 ) và ( 2 )
=> \(\frac{2}{5}< S< \frac{8}{9}\)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
Chứng minh rằng \(\frac{2}{5}\)< S <\(\frac{8}{9}\)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\left(1\right)\)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}
bạn ơi dòng đầu tiên bạn tách sai rồi theo minh thì không phải thế đâu
\(S=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\frac{1}{8\cdot8}+\frac{1}{9\cdot9}\)
HÃY CHỨNG MINH \(\frac{2}{5}< S< \frac{7}{8}\)
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
dạng 1 : so sánh
a) P = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}+\frac{1}{2014^2}\)và Q = \(1\frac{3}{4}\)
dạng 2 : toán chứng minh
1. cho S = \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{130}\)chứng minh rằng : \(\frac{1}{4}< S< \frac{91}{330}\)
2. cho S = \(\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+...+\frac{5}{49}\). CMR : 3 < S < 8
3. CMR : \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{1999}}>1000\)
2.a) Vào question 126036
b) Vào question 68660