\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2015.2016}\)

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

\(S=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)

làm r sao cứ đăng hoài vậy?

S=1/1-1//2+1/2-1/3+1/3-1/4+.......=1/2014-1/2015

S=1/1-1/2015

S=2015/2015-1/2015

S=2014/2015

ta có: \(\frac{1}{1.2}>0;\frac{1}{2.3}>0;...;\frac{1}{n.\left(n+1\right)}>0\)

\(\Rightarrow S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n.\left(n+1\right)}>0\)

ta có: \(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n.\left(n+1\right)}\)

\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(S=1-\frac{1}{n+1}< 1\)

=> 0 < S < 1

=> S không phải là số tự nhiên

cau hỏi tương tự ko có mà!!!!!!!!!!!!!!!!!!!!!!!!!!!!

3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)

3C=2014.2015.2016

C=2014.2015.2016:3

TẠi sao lại có số 1 ở đầu vậy?    

 Đặt A = 1 + 2 + 3 + 4 + ....... + 2014

Số các số hạng của A là:

2014 - 1  + 1 = 2014 (số)

A = 2014.(2014 + 1):2 = 2029105

Đặt B = 1.2 + 2.3 + 3. 4 + .....+ 2014.2015

3B = 1.2.3 + 2.3.3 + .3.4.3 + ......... + 2014.2015

3B = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + .......... + 2014.2015.(2016 - 2013)

3B = 1.2.3 + 2.3.4 - 1.2.3 + ........+2014.2015.2016 - 2013.2014.2015

3B = 2014 . 2015 .2016 = 8181351360

B = 8181351360 : 3

B = 2727117120

Vậy D = A + B = 2029105 + 2727117120 = 2729146225

Đk: n khác 0, n khác -1

\(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{n\left(n+1\right)}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)

\(=1-\dfrac{1}{n+1}\)

Vì \(0< \dfrac{1}{n+1}< 1\) (n khác 0, n khác -1) nên \(0< 1-\dfrac{1}{n+1}< 1\)

hay 0<S<1

Vậy S không là stn

A= 1-1/2015=2014/2015

=> 2014/2015+1/2015=2x

=> 1=2x

=> x=1/2

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}< 1\) (đpcm)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(=\frac{49}{50}\)

\(\Rightarrow\) Quy đồng phân số và 1 là : \(\frac{49}{50}\) và \(1\)

Giữ nguyên phân số \(\frac{49}{50}\)

Ta có : \(\frac{1}{1}=\frac{1.50}{1.50}=\frac{50}{50}\)

\(\Rightarrow\frac{49}{50}< \frac{50}{50}\left(đpcm\right)\)

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: \(\frac{1}{1.2+2.3+3.4}=\frac{3}{3.4.5}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(=1-\frac{1}{2015}\)

\(=\frac{2014}{2015}\)

A=1/1-1/2+1/2-1/3+1/3-...........+1/2014-1/2015

A=1/1-1/2015

A=2014/2015