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27 tháng 3 2018

\(S=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2n^2}=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)

Lại có: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)

=> \(S=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2n^2}=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}.1=\frac{1}{4}\)

=> \(S< \frac{1}{4}\)

20 tháng 4 2018

Ta có :

\(S=\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+.......+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}\)

\(\Rightarrow S< \frac{1}{17}+\frac{1}{17}+......+\frac{1}{17}+\frac{1}{17}+\frac{1}{17}\)

\(\Rightarrow S< \frac{1}{17}.48\)

\(\Rightarrow S< \frac{48}{17}\)

\(\Rightarrow S< 2\)( 1 ) 

Lại có :

\(S>\frac{1}{64}+\frac{1}{64}+.........+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}\)

\(\Rightarrow S>\frac{1}{64}.48\)

\(\Rightarrow S>\frac{3}{4}\)( 2 ) 

Từ ( 1 ) và ( 2 ) suy ra : \(\frac{3}{4}< S< 2\)

Vậy \(1< S< 2\left(ĐPCM\right)\)

a)\(\dfrac{1}{2^2}<\dfrac{1}{1.2}\)

\(\dfrac{1}{3^3}<\dfrac{1}{2.3}\)

\(...\)

\(\dfrac{1}{8^2}<\dfrac{1}{7.8}\)

Vậy ta có biểu thức:

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)

\(B= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(B<1-\dfrac{1}{8}=\dfrac{7}{8}<1\)

Vậy B < 1 (đpcm)

 

 

 

Giải:

a) Ta có:

1/22=1/2.2 < 1/1.2

1/32=1/3.3 < 1/2.3

1/42=1/4.4 < 1/3.4

1/52=1/5.5 < 1/4.5

1/62=1/6.6 < 1/5.6

1/72=1/7.7 < 1/6.7

1/82=1/8.8 <1/7.8

⇒B<1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8

   B<1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8

   B<1/1-1/8

   B<7/8

mà 7/8<1

⇒B<7/8<1

⇒B<1

b)S=3/1.4+3/4.7+3/7.10+...+3/40.43+3/43.46

   S=1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

   S=1/1-1/46

   S=45/46

Vì 45/46<1 nên S<1

Vậy S<1

Chúc bạn học tốt!

13 tháng 4 2015

bài này có trông sách nâng cao và phataienf toán 6ss tr

4 tháng 5 2017

Ta có :

\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

\(S=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Nhận xét :

\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{2}\rightarrowđpcm\)

22 tháng 6 2015

Ta có:

\(\frac{1}{5}=\frac{1}{5}\)

\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}<\frac{1}{12}.3=\frac{1}{4}\)

\(\frac{1}{61}+\frac{1}{62}<\frac{1}{60}.2=\frac{1}{30}\)

=>\(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}<\frac{1}{5}+\frac{1}{4}+\frac{1}{30}\)

=>\(S<\frac{29}{60}<\frac{30}{60}=\frac{1}{2}\)

=>\(S<\frac{1}{2}\)(đpcm)

20 tháng 6 2015

Ta có: 

\(\frac{1}{5}=\frac{1}{5}\)

\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}<\frac{1}{12}.3=\frac{1}{4}\)

\(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}<\frac{1}{60}.3=\frac{1}{20}\)

=>S<\(\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)

=>\(S<\frac{1}{20}\)(đpcm)

20 tháng 6 2015

Ta có: \(S=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)<\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{13}+\frac{1}{13}\right)+\left(\frac{1}{61}+\frac{1}{61}+\frac{1}{61}\right)\)\(\Rightarrow S<\frac{1}{5}+\frac{3}{13}+\frac{3}{61}<\frac{1}{5}+\frac{3}{12}+\frac{3}{60}=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)