= 1 - 1/4 +1/4 -1/7 + 1/7 -1/10+....+ 1/n-1/n+3

= 1- 1/n+3 (<1)

Ta có 

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)

\(S=1-\frac{1}{n+3}< 1\)(vì n thuộc N*)

_Kudo_

Cảm ơn bn

Bài làm

S= 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/n. (n+3)

  =1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... +1/(n-3) - 1/n +1/n - 1/(n+3)

  =1/1 + ( - 1/4 + 1/4 - 1/7 + 1/7 - ... -1/n + 1/n ) -1/ (n+3)

  = 1 + 0 - 1/(n+3)

  = 1 - 1/(n+3)

Mà 1 - 1/(n+3) < 1

Vậy S < 1

Mk trình bày ko đc chi tiết lắm ,sorry bạn nha

sorry mình cũng đang muốn hỏi bài nay

\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3n}{n\left(n+3\right)}\)

\(S=\frac{4-1}{1\cdot4}+\frac{7-4}{4\cdot7}+\frac{10-7}{7\cdot10}+...+\frac{n+3-3}{n\left(n+3\right)}\)

\(S=\frac{4}{1\cdot4}-\frac{1}{1\cdot4}+\frac{7}{4\cdot7}-\frac{4}{4\cdot7}+...+\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n-3}\)

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

bé hơn 1 chứ ko lớn hơn 1 đc đâu

cute phô mai que chắc mình nhầm đề. Thanks bn nha <3

\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{n\left(n+3\right)}\)

\(\Rightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n-3}\)

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

\(\Rightarrow S=\frac{n+2}{n+3}\)

\(\Rightarrow S< 1\)

A=\(\frac{3}{1.4}+\frac{3}{4.7}+...........+\frac{3}{n.\left(n+3\right)}\)

A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...............+\frac{1}{n}-\frac{1}{n+3}\)

A=\(1-\frac{1}{n+3}\)<1

Vậy A<1(đpcm)

Do : \(\frac{3}{1.4}=\frac{1}{1}-\frac{1}{4};\frac{3}{4.7}=\frac{1}{4}-\frac{1}{7}\).... tuong tu ... \(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)

S= \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n-3}-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+3}\)

S= \(1-\frac{1}{n+3}\)<1

=> S<1 (dpcm)

(do : 3/ 1.4 = 1/1 - 1/4;  3/4.7= 1/4 - 1/7 ...

S= 1- 1/4 + 1/4 + 1/4 - 1/7 + ... + 1/ n - 1/ (n+3)

S= 1- 1/ (n+3) <1 

=> S <1 (dpcm)

ta có S = 1-1/4 + 1/4 - 1/7 =....................................+1/n - 1/(n+1) = 1- 1/(n+1)

 mà n thuộc N* nên S<1

ta có \(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{\left(n+3\right)}\)

\(S=1-\frac{1}{\left(n+3\right)}\)

thì đương nhiên S nhỏ hơn 1 rồi

=>S= 1- 1/4 + 1/4 -1/7 + 1/7 - 1/10 +...+ 1/n - 1/(n+3)

=>S= 1- 1/(n+3)

=>S + 1/(n+3) = 1

=>S<1