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\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}=\frac{5}{6}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(B=1-\frac{1}{n+1}=\frac{n}{n+1}\)
\(P=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+.....+\frac{4033}{\left(2016.2017\right)^2}\)
\(\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+.....+\frac{2017^2-2016^2}{2016^2.2017^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+....+\frac{1}{2016^2}-\frac{1}{2017^2}\)
\(=1-\frac{1}{2017^2}< 11\) (đpcm)
Bài này trong đề thi học kì 2 môn Toán lớp 6 trường Amsterdam năm 2016-2017 này. Mình 10 luôn hehe
\(3D_n=1.2.3+2.3.3+3.4.3+...+n\left(n+1\right)3\)
\(=1.2\left(3-0\right)+2.3\left(4-1\right)+...+n\left(n+1\right)\left(n+2-n+1\right)\)
\(=1.2.3-0.1.2+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(=n\left(n+1\right)\left(n+2\right)-0.1.2=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow D_n=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Dn = 1.2 + 2.3 + 3.4 +...+ n(n + 1)
3Dn = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) +...+ n(n + 1).[(n + 2) - (n - 1)]
3Dn = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +...+ n(n + 1)(n + 2) - (n - 1)n(n + 1)
3Dn = [1.2.3 + 2.3.4 + 3.4.5 +...+ n(n + 1)(n + 2)] - [0.1.2 + 1.2.3 + 2.3.4 +...+ n(n - 1)(n + 1)]
3Dn = n(n + 1)(n + 2) - 0.1.2
3Dn = n(n + 1)(n + 2)
Dn = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\) (đpcm)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}< 1\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}< 1\)
\(S=1-\frac{1}{50}< 1\)
\(S=\frac{49}{50}< 1\left(đpcm\right)\)
1 : dễ mà
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)
1 phần 1 - 1 phần 2 = 1 phần 1.2 mà tương tự như thế đó
=> 1 - 1 phần n+1
đS
\(\frac{1}{1.2}+\frac{1}{2.3}+..........+\frac{1}{n.\left(n+1\right)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+............+\frac{1}{n}-\frac{1}{n+1}\)
\(=1-\frac{1}{n+1}\)
\(=\frac{n}{n+1}\)
Bài 2:Ta có:\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};.................;\frac{1}{n^2}<\frac{1}{\left(n-1\right).n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...........+\frac{1}{n^2}<\frac{1}{1.2}+\frac{1}{2.3}+.........+\frac{1}{\left(n-1\right).n}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...........+\frac{1}{n-1}-\frac{1}{n}\)
=\(1-\frac{1}{n}<1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...........+\frac{1}{n^2}<1\)
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) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)
\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)
\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}.\frac{2013}{4030}\)
\(=\frac{6039}{8060}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{4}{5}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{4}{5}\)
\(1-\frac{1}{x+1}=\frac{4}{5}\)
\(\frac{x}{x+1}=\frac{4}{5}\)
\(\frac{x}{x+1}=\frac{4}{4+1}\)
\(\Rightarrow x=4\)
Vậy x = 4
=))
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{4}{5}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{4}{5}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{4}{5}\)
\(\Leftrightarrow\frac{1}{x-1}=1-\frac{4}{5}\)
\(\Leftrightarrow\frac{1}{x-1}=\frac{1}{5}\)
\(\Leftrightarrow x-1=5\)
\(\Leftrightarrow x=5+1\)
\(\Leftrightarrow x=6\)
~ Rất vui vì giúp đc bn ~ ^_<
A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
Ta có: \(\frac{1}{1.2}=\frac{3}{1.2.3}\) ;\(\frac{1}{1.2+2.3}=\frac{3}{2.3.4}\); \(\frac{1}{2.3+3.4}=\frac{3}{3.4.5}\); ......;\(\frac{1}{1.2+2.3+3.4+...+n\left(n+1\right)}=\frac{3}{n\left(n+1\right)\left(n+2\right)}\)
=> \(S=\frac{3}{1.2.3}+\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
Ta lại có: \(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\); \(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\); \(\frac{2}{3.4.5}=\frac{1}{3.4}-\frac{1}{4.5}\);....;\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)=> \(S=\frac{3}{4}-\frac{3}{2\left(n+1\right)\left(n+2\right)}< \frac{3}{4}\)
=> \(S< \frac{3}{4}\)
Mình nhầm 1 chỗ: \(\frac{1}{1.2+2.3+3.4}=\frac{3}{3.4.5}\)