=>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

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

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

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

Ta có :

\(\frac{1}{n+3}>0\)

\(\Leftrightarrow-\frac{1}{n+3}< 0\)

\(\Leftrightarrow1-\frac{1}{n+3}< 1\)

\(\Leftrightarrow S< 1\left(đpcm\right)\)

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n.\left(n+3\right)}\)

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

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

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

Vi \(n\inℕ^∗\)nên \(n+2< n+3\)

DO đó\(\frac{n+2}{n+3}< 1\)

Vậy S <1

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

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

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

     \(=1-\frac{1}{n+3}<1\)

\(\Rightarrow\) S < 1 ( đpcm )

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

=> S = 1 - \(\frac{1}{n+3}\)

vậy S = 1-  \(\frac{1}{n+3}\)

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\)với \(n\in N\ne0\)

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

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

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

\(\Rightarrow S< 1\left(ĐPCM\right)\)

S=1/1-1/4+1/4-1/7+.........+1/N-1/N+1

=1/1-(1/4-1/4)+...............+(1/N-1/N)-1/N+1

=1-1/N+1

->S<1

NHA!

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\)

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

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

Vậy S<1 (đpcm)

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

=> S = 1 - \(\frac{1}{n+3}\)

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

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

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

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

P/s: Đến đó thôi.......^.^

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

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

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

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

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

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

Câu 1 dễ thôi. Bạn tính tử, rồi tính mẫu sao cho khi phân phối ở cả tử và mẫu đều có phần thừa số có thể rút gọn cho nhau. Giờ mik bận quá nên ko thể giải dầy đủ. Thông cảm nha...

Câu 2: Cũng ko khó lắm đâu:

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

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

Vậy: S<1

Để làm dc bài sau, bạn nhớ giùm mik công thức: \(\frac{a}{b.\left(b+a\right)}\)=\(\frac{1}{b}\)-\(\frac{1}{b+a}\)

Câu 3:  Đặt \(A=\frac{2003.2004-1}{2003.2004}\), \(B=\frac{2004.2005-1}{2004.2005}\)ta có:

\(A=\frac{2003.2004}{2003.2004}\)-\(\frac{1}{2003.2004}\)=1-\(\frac{1}{2003.2004}\)

\(B=\frac{2004.2005}{2004.2005}\)-\(\frac{1}{2004.2005}\)=1-\(\frac{1}{2004.2005}\)

Vì 2003.2004<2004.2005=>\(\frac{1}{2003.2004}\)>\(\frac{1}{2004.2005}\)

=>1-\(\frac{1}{2003.2004}\)<1-\(\frac{1}{2004.2005}\)

Vậy:  \(\frac{2003.2004-1}{2003.2004}\)< \(\frac{2004.2005-1}{2004.2005}\)

Nhớ cho mik nha. Thanks

\(\frac{1}{1.3}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{n\left(n+3\right)}=\frac{2018}{6057}\)

\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\right)=\frac{2018}{6057}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}=\frac{2018}{6057}.3\)

\(\Rightarrow1-\frac{1}{n+3}=\frac{2018}{2019}\)

\(\Rightarrow\frac{1}{n+3}=1-\frac{2018}{2019}\)

\(\Rightarrow\frac{1}{n+3}=\frac{1}{2019}\)

\(\Rightarrow n+3=2019\)

\(\Rightarrow n=2016\)

Vậy n = 2016