Bài 1 :Ký hiệu n ! = 1.2.3.4.....(n-1).n
chứng minh rằng:A = 1/2!+2/3!+3/4!+.....+2015/2016! < 1
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ta có:1/2!<1
2/3!<1
......
......
2015/2016!<1
=>A=1/2!+2/3!+3/4!+......+2015/2016! luôn luôn <1
câu 1: tích 1.2.3.4...2015 hơn tích 1.2.3.4...2014 1 thừa số là thừa số 2014
=[1.2.3.4...(2014.2014)]-1.2.3.4...20142
=> tích đó =0
câu 2:
2016x +(1+3+5+ …+2015) = 2016 (*)
Tổng : 1+3+5+ …+2015 có: (2015-1):2+1= 1008 số hạng
= > Tổng : 1+3+5+ …+2015 có: 504 cặp số
Từ (*) = > 1009x + (2015+1).504 = 2016
= > 1009x = 2016.(1-504) = > x = (-1006)
câu 2 sai vì tui nhìn tưởng đề là x+(x+1)+(x+3)+(x+5)+...+(x+2015)=2016
\(N=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(N=\left(1+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+...+\frac{1}{2016}\right)\)
\(N=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1008}\right)\)
\(N=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}=K\)
\(E=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{2015}{3^{2015}}-\dfrac{2016}{3^{2016}}\\ 3E=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{2015}{3^{2014}}-\dfrac{2016}{3^{2015}}\\ 3E+E=\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{2015}{3^{2014}}-\dfrac{2016}{3^{2015}}\right)+\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{2015}{3^{2015}}-\dfrac{2016}{3^{2016}}\right)\\ 4E=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}-\dfrac{2016}{3^{2016}}\\ 4E< 1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}\left(1\right)\)
Gọi \(D=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{2015}}\)
\(3D=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{2013}}-\dfrac{1}{3^{2014}}\\ 3D+D=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{2013}}-\dfrac{1}{3^{2014}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}\right)\\ 4D=3-\dfrac{1}{3^{2015}}< 3\\ \Rightarrow D< \dfrac{3}{4}\left(2\right)\)
Từ (1) và (2) ta có:
\(4E< \dfrac{3}{4}\\ \Rightarrow E< \dfrac{3}{16}\)
Bài 3
\(\frac{n+6}{n+1}=\frac{n+1+5}{n+1}=\frac{n+1}{n+1}+\frac{5}{n+1}\)
\(=1+\frac{5}{n+1}\)
Vậy để \(\frac{n+6}{n+1}\in Z\Rightarrow1+\frac{5}{n+1}\in Z\)
Hay \(\frac{5}{n+1}\in Z\)\(\Rightarrow n+1\inƯ_5\)
\(Ư_5=\left\{1;-1;5;-5\right\}\)
* \(n+1=1\Rightarrow n=0\)
* \(n+1=-1\Rightarrow n=-2\)
* \(n+1=5\Rightarrow n=4\)
* \(n+1=-5\Rightarrow n=-6\)
Vậy \(n\in\left\{0;-2;4;-6\right\}\)
Bài 2:
\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right)\\ =2\left(\frac{1}{3}-\frac{1}{28}\right)\\ =2.\frac{56}{84}\\ =\frac{56}{42}=\frac{28}{21}\)