Tính
1/3+1/6+1/10+...+2/n(n+1)=2015/2016
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TÍnh S=3/1.4+3/4.7+3?7>!0+...+3/n(n+3) với n là số tự nhiên . chứng minh S<1
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:
\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right)\\ =2\left(\frac{1}{3}-\frac{1}{28}\right)\\ =2.\frac{56}{84}\\ =\frac{56}{42}=\frac{28}{21}\)
\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{n.\left(n+1\right)}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{n.\left(n+1\right)}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}...+\frac{1}{n.\left(n+1\right)}=\frac{2015}{2016}\)
\(\frac{1.2}{3.2}+\frac{1.2}{6.2}+\frac{1.2}{10.2}+...+\frac{1}{n\left(n+1\right)}=\frac{2015}{2016}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{1}{n.\left(n+1\right)}=\frac{2015}{2016}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{2015}{2016}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2015}{2016}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{2015}{2016}\)
\(2.\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2015}{2016}\)
\(\frac{1}{2}-\frac{1}{n+1}=\frac{2015}{2016}:2\)
\(\frac{1}{2}-\frac{1}{n+1}=\frac{2015}{4032}\)
\(\frac{1}{n+1}=\frac{1}{2}-\frac{2015}{4032}\)
\(\frac{1}{n+1}=\frac{1}{4032}\)
\(\Rightarrow n+1=4032\)
\(\Rightarrow n=4031\)
=2015-(2015-2016)-2016+22017-2015-22015/22014-(1-4)-3-(5+6)+11
=(2015-2015)+(2016-2016)+22-2+3-3-11+11
=0+0+(4-2)+(3-3)-(11-11)
=2
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{2015}{2016}\)
\(\Leftrightarrow2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{2015}{2016}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{2015}{2016}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{2015}{4032}\)
\(\Leftrightarrow\dfrac{1}{n+2}=\dfrac{1}{4032}\)
hay n=4030