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:

\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right)\\ =2\left(\frac{1}{3}-\frac{1}{28}\right)\\ =2.\frac{56}{84}\\ =\frac{56}{42}=\frac{28}{21}\)

\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{n.\left(n+1\right)}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{n.\left(n+1\right)}\)

còn lại tự làm

mình giải nhé:

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{2015}{2016}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{n\left(n+1\right)}=\frac{2015}{2016}\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{n\left(n+1\right)}=\frac{2015}{2016}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2015}{2016}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{2015}{2016}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2015}{2016}\)

\(\Leftrightarrow1-\frac{2}{n+1}=\frac{2015}{2016}\Leftrightarrow\frac{2}{n+1}=\frac{1}{2016}\)

\(\Leftrightarrow n+1=4032\Leftrightarrow n=4031\)

Xong rồi nhé ^ ^ chúc bạn học tốt!

chị kết bạn với em nha gửi lời kết bn với em nhé

j zậy em hả 

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}...+\frac{1}{n.\left(n+1\right)}=\frac{2015}{2016}\)

\(\frac{1.2}{3.2}+\frac{1.2}{6.2}+\frac{1.2}{10.2}+...+\frac{1}{n\left(n+1\right)}=\frac{2015}{2016}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{1}{n.\left(n+1\right)}=\frac{2015}{2016}\)

\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{2015}{2016}\)

\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2015}{2016}\)

\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{2015}{2016}\)

\(2.\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2015}{2016}\)

\(\frac{1}{2}-\frac{1}{n+1}=\frac{2015}{2016}:2\)

\(\frac{1}{2}-\frac{1}{n+1}=\frac{2015}{4032}\)

\(\frac{1}{n+1}=\frac{1}{2}-\frac{2015}{4032}\)

\(\frac{1}{n+1}=\frac{1}{4032}\)

\(\Rightarrow n+1=4032\)

\(\Rightarrow n=4031\)

=2015-(2015-2016)-2016+2^2017-2015-2²⁰¹⁵/2²⁰¹⁴-(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

giải thích củ thể dùm mình cái