Chứng tỏ rằng:
a) 1 1.2 + 1 2.3 + ... + 1 9.10 < 1
b) 3 2.5 + 3 5.8 + 3 8.11 + 3 11.14 + 3 14.17 + 3 17.20 < 1 2
c) 1 2 2 + 1 3 2 + 1 4 2 + ... + 1 25 2 < 1
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a) = 1-1/2+1/2-1/3+1/3-1/4
= 1-1/4=3/4
b)=1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017+1/2017-1/2018
=1-1/2018=2017/2018
c)=1/2-1/5+1/5-1/8+1/8-1/11+1/2009-1/2012+1/2012-1/2015
= 1/2-1/2015=2015/4030-2/4030=2013/4030
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\)
b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017-2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
c) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)
\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)
\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}.\frac{2013}{4030}\)
\(=\frac{6039}{8060}\)
Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{50-49}{49.50}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A=1-\frac{1}{50}=\frac{49}{50}\)
B=\(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{14.17}+\frac{3}{17.20}\)
\(\Rightarrow B=\frac{5-2}{2.5}+\frac{8-5}{5.8}+...+\frac{17-14}{14.17}+\frac{20-17}{17.20}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{20}=\frac{10}{20}-\frac{1}{20}=\frac{9}{20}\)
Cau a) 1/1.2 +1/2.3 +1/3.4+...+1/99.100= 1/1-1/2+1/2-1/3+...+1/99-1/100
=1/1-1/100=99/100
99/100<1 thì 1/1.2 +1/2.3+1/3.4+...+1/99.100<1
Câu b): Ta có: 1/2^2<1/1.2
1/3^2<1/2.3
...............(so sánh như vậy với các số khác)
1/2016^2<1/2015.2016
Áp dụng của câu a ta thêm vào sau về thành: 1/1.2+1/2.3+1/3.4+...+1/2015.2016
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2015-1/2016
=1/1-1/2016
=2015/2016<1
Ma :1/2^2+1/3^2+1/4^2+...+1/2016^2<1/1.1+1/2.3+1/3.4+...+1/2015.2016
Nen:1/1^2+1/3^2+1/4^2+...+1/2016^2<1
a) Vì n.(n+1) = 1/n-1/n+1 suy ra n thuộc N n khác 0
b) A=1/1*2+1/2*3+1/3*4+...+1/9.10
A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10
A=1-1/10=9/10
Vậy A = 9/10
a) 1/1.2 + 1/2.3 + 1/3.4 + ....... + 1/99.100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..... + 1/99 - 1/100
= 1 - 1/100
= 99/100 < 1 nên 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/99.100 < 1 (ĐPCM)
a)1-1/2+1/2-1/3+1/3-1/4+......+1/99-1/100
1-1/100=99/100<1
cho mk nha ^^
a: \(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)
b: \(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{9}{10}\)
a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{44}{45}\)
=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{45}\)
=> \(1-\frac{1}{x+1}=\frac{44}{45}\)
=> \(\frac{x}{x+1}=\frac{44}{45}\)
=> x = 44
b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
.................
\(\frac{1}{45^2}< \frac{1}{44.45}=\frac{1}{44}-\frac{1}{45}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{44}-\frac{1}{45}=1-\frac{1}{45}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)
a) 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/(x+1)=1-1/(x+1)=x/(x+1)=44/45
=> x=44
b/ 1/22 < 1/1.2; 1/32 < 1/2.3; ....; 1/452 < 1/44.45
=> A < 1/1.2+1/2.3+...+1/44.45=1-1/45=44/45 < 1
=> A < 1
Phần chứng tỏ quy đồng lên rồi tính là ra
Còn phần tính S thì áp dụng tính chất vừa chứng tỏ để tách ra
Kết quả là 49/50
Ta có: \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)
\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)\(< \frac{17}{34}=\frac{1}{2}\)
\(\Rightarrow\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{14.17}< \frac{1}{2}\)
Vậy:..........................................(đpcm)
a ) 1 1.2 + 1 2.3 + ... + 1 9.10 = 9 10 < 1.
b ) 3 2.5 + 3 5.8 + 3 8.11 + 3 11.14 + 3 14.17 + 3 17.20 = 1 2 − 1 20 < 1 2 .
c ) T a c ó 1 2 2 < 1 1.2 ; 1 3 2 < 1 2.3 ; 1 4 2 < 1 3.4 ; ... ; 1 25 2 < 1 24.25 .
D o đ ó , 1 2 2 + 1 3 2 + 1 4 2 + 1 25 2 < 1 1.2 + 1 2.3 + 1 3.4 + ... + 1 24.25 < 1.