\(Cho:S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}n\in Nsao\)
Chứng minh : S<1
K
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Những câu hỏi liên quan
H
12 tháng 4 2019
Ta có:
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n.\left(n+3\right)}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{n+3}\)
\(\Leftrightarrow S=\frac{n+3}{n+3}-\frac{1}{n+3}=\frac{n+3-1}{n+3}=\frac{n+2}{n+3}\)
\(\Rightarrow\frac{n+2}{n+3}< 1\Rightarrow S< 1\)
9 tháng 1 2015
- 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 = \(1-\frac{1}{n+3}\)
\(\Rightarrow\) S < 1 ( đpcm )
9 tháng 4 2017
=> 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- \(\frac{1}{n+3}\)
9 tháng 4 2015
=>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
28 tháng 2 2018
\(S=\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{n\left(n+3\right)}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{n}-\frac{1}{n+3}\)
\(=1-\frac{1}{n+3}\)
Ta có :
\(\frac{1}{n+3}>0\)
\(\Leftrightarrow-\frac{1}{n+3}< 0\)
\(\Leftrightarrow1-\frac{1}{n+3}< 1\)
\(\Leftrightarrow S< 1\left(đpcm\right)\)
28 tháng 2 2018
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n.\left(n+3\right)}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(S=1-\frac{1}{n+3}\)
\(S=\frac{n+2}{n+3}\)
Vi \(n\inℕ^∗\)nên \(n+2< n+3\)
DO đó\(\frac{n+2}{n+3}< 1\)
Vậy S <1
16 tháng 7 2016
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!
WH
5 tháng 3 2018
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(\Rightarrow S=1-\frac{1}{n+3}\)
\(\Rightarrow S=\frac{n+3-1}{n+3}\)
\(\Rightarrow S=\frac{n+2}{n+3}\)
P/s: Đến đó thôi.......^.^
S
5 tháng 3 2018
\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+....+\frac{3}{n\cdot\left(n+3\right)}\)
\(S=\frac{4-1}{1\cdot4}+\frac{7-4}{4\cdot7}+\frac{10-7}{7\cdot10}+....+\frac{\left(n+3\right)-n}{n\cdot\left(n+3\right)}\)
\(S=\left(\frac{4}{1\cdot4}-\frac{1}{1\cdot4}\right)+\left(\frac{7}{4\cdot7}-\frac{4}{4\cdot7}\right)+\left(\frac{10}{7\cdot10}-\frac{7}{7\cdot10}\right)+.....+\left(\frac{n+3}{n\cdot\left(n+3\right)}-\frac{n}{n\cdot\left(n+3\right)}\right)\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{n}-\frac{1}{n+3}\)
\(S=1-\frac{1}{n+3}\)
\(S=\frac{n+3}{n+3}-\frac{1}{n+3}=\frac{n+2}{n+3}\)
14 tháng 7 2018
Ta có :
3/ 1.4 + 3/ 4.7 + 3/ 7.10 + ... + 3/ n( n + 1 )
= 1 - 1/4 + 1/4 - 1/7 + ... + 1/ n - 1/ n + 3 .
= 1 - 1/ n+3 .
= n+3 - 1 / n+3
= n+2 / n+3 .
9 tháng 11 2016
\(\frac{1}{1.3}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{n\left(n+3\right)}=\frac{2018}{6057}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\right)=\frac{2018}{6057}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}=\frac{2018}{6057}.3\)
\(\Rightarrow1-\frac{1}{n+3}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{n+3}=1-\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{n+3}=\frac{1}{2019}\)
\(\Rightarrow n+3=2019\)
\(\Rightarrow n=2016\)
Vậy n = 2016
\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{n}-\frac{1}{n+3}\)
\(=1-\frac{1}{n+3}<1\)