1/1.4 + 1/4.7 + 1/7.11 + ... + 1/n(n + 3) = 89/270
tìm n :
\(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{n\left(n+3\right)}=\frac{89}{270}\)
tìm số tự nhiên n biết\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{n\left(n+3\right)}=\frac{2018}{6057}\)
\(\frac{1}{1.3}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{n\left(n+3\right)}=\frac{2018}{6057}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\right)=\frac{2018}{6057}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}=\frac{2018}{6057}.3\)
\(\Rightarrow1-\frac{1}{n+3}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{n+3}=1-\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{n+3}=\frac{1}{2019}\)
\(\Rightarrow n+3=2019\)
\(\Rightarrow n=2016\)
Vậy n = 2016
Cho S=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{n.\left(n+3\right)}\) với n thuộc N*
Chứng tỏ rằng S<1
=>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
Cho S = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n.\left(n+3\right)}\) vs n thuộc N*. Chứng minh S < 1
Ta có:
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n.\left(n+3\right)}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{n+3}\)
\(\Leftrightarrow S=\frac{n+3}{n+3}-\frac{1}{n+3}=\frac{n+3-1}{n+3}=\frac{n+2}{n+3}\)
\(\Rightarrow\frac{n+2}{n+3}< 1\Rightarrow S< 1\)
\(Cho:S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}n\in Nsao\)
Chứng minh : S<1
\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{n}-\frac{1}{n+3}\)
\(=1-\frac{1}{n+3}<1\)
a) Tính \(\frac{1}{n}-\frac{1}{n+1}\left(n\inℕ^∗\right)\)
b)Tính \(\frac{1}{n}-\frac{1}{n+3}\left(n\inℕ^∗\right)\)
c) Tính \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{3}{91.94}\)
d) Tính \(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{91.94}\)
c)1*(1/2-1/3+1/3-1/4+.....+1/91-1/94)
1/2-1/94 ban tu tinh nhe
d)1*(1/1-1/4+1/4-1/7+......+1/91-1/94)
1-1/94 ban tu tinh nhe
tk nha
a) \(\frac{1}{n}-\frac{1}{n+1}\left(n\inℕ^∗\right)\)
\(\Leftrightarrow\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}\Leftrightarrow\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
b) \(\frac{1}{n}-\frac{1}{n+3}\left(n\inℕ^∗\right)\)
\(\Leftrightarrow\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}=\frac{n+3-n}{n\left(n+3\right)}=\frac{3}{n\left(n+3\right)}\)
c,d dễ bn tách ra rồi trừ đi
\(A.\)\(\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n}+\frac{-1}{n+1}=\frac{n+1}{n\left(n+1\right)}+\frac{-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
\(B.\)\(\frac{1}{n}-\frac{1}{n+3}=\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}=\frac{3}{n\left(n+3\right)}\)
\(C.\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(D.\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{91.94}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{94}=\frac{1}{1}-\frac{1}{94}=\frac{93}{94}\)
Cho s = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\)n \(\in\)N KHÁC 0. Chứng minh s < 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 \(n\in N\ne0\)
\(\Rightarrow S=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{10}\right)+...+\left(\frac{1}{n}-\frac{1}{n+3}\right)\)
\(\Rightarrow S=1-\frac{1}{4}+\frac{1}{7}-\frac{1}{7}+\frac{1}{10}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(\Rightarrow S=1-\frac{1}{n+3}\)
\(\Rightarrow S< 1\left(ĐPCM\right)\)
a) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\frac{125}{376}\left(x€N\cdot\right)\)
b) \(\frac{3}{4}x-14\frac{2}{3}:\left(\frac{11}{15}+\frac{1111}{3535}+\frac{111111}{636363}\right)=12\)
1 rút gọn :
A= \(\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
2 cho s= \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.......+\frac{3}{n\left(n+3\right)}\)n thuộc N chứng minh s<1
Cho S = \(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+\(\frac{3}{7.10}\)+ . . . + \(\frac{3}{n\left(n+3\right)}\)( n thuộc N* )
Chứng minh 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
\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(=1-\frac{1}{n+3}< 1\)