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 = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n.\left(n+3\right)}\) 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=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{n.\left(n+3\right)}\) 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
\(Cho:S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}n\in Nsao\)
Chứng minh : S<1
\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{n}-\frac{1}{n+3}\)
\(=1-\frac{1}{n+3}<1\)
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 \(n\in N\ne0\)
\(\Rightarrow S=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{10}\right)+...+\left(\frac{1}{n}-\frac{1}{n+3}\right)\)
\(\Rightarrow S=1-\frac{1}{4}+\frac{1}{7}-\frac{1}{7}+\frac{1}{10}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(\Rightarrow S=1-\frac{1}{n+3}\)
\(\Rightarrow S< 1\left(ĐPCM\right)\)
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
S=\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+\(\frac{3}{7.10}\)+......+\(\frac{3}{n.\left(n+3\right)}\)với n thuộc N*
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(\Rightarrow S=1-\frac{1}{n+3}\)
\(\Rightarrow S=\frac{n+3-1}{n+3}\)
\(\Rightarrow S=\frac{n+2}{n+3}\)
P/s: Đến đó thôi.......^.^
\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+....+\frac{3}{n\cdot\left(n+3\right)}\)
\(S=\frac{4-1}{1\cdot4}+\frac{7-4}{4\cdot7}+\frac{10-7}{7\cdot10}+....+\frac{\left(n+3\right)-n}{n\cdot\left(n+3\right)}\)
\(S=\left(\frac{4}{1\cdot4}-\frac{1}{1\cdot4}\right)+\left(\frac{7}{4\cdot7}-\frac{4}{4\cdot7}\right)+\left(\frac{10}{7\cdot10}-\frac{7}{7\cdot10}\right)+.....+\left(\frac{n+3}{n\cdot\left(n+3\right)}-\frac{n}{n\cdot\left(n+3\right)}\right)\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{n}-\frac{1}{n+3}\)
\(S=1-\frac{1}{n+3}\)
\(S=\frac{n+3}{n+3}-\frac{1}{n+3}=\frac{n+2}{n+3}\)
Cho S=\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+\(\frac{3}{7.10}\)+........+\(\frac{3}{n\left(n+3\right)}\) với e n*
Chứng minh rằng S<1
giúp mk nha , mk đang cần gấp!!! Thank nhìu!!!! ^_^
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 - \(\frac{1}{n+3}\)
Tính tổng
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\)
=1/1-1/4+1/4-1/7+....+1/n-1/n+3
=1-1/n+3
=n+2/n+3
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 .
\(3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{n.\left(n+3\right)}\right)\)
\(3.\left(1.\frac{1}{4}+\frac{1}{4}.\frac{1}{7}+\frac{1}{7}.\frac{1}{10}+...+\frac{1}{n}.\frac{1}{n+3}\right)\)
\(3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\right)\)
\(3.\left(1-\frac{1}{n+3}\right)=3.\frac{n+2}{n+3}=\frac{6n+6}{n+3}\)
Chào mọi người, cũng khá lâu rồi mình không vào olm, chắc cũng vài tháng rồi. Giải giúp mình bài này với... Mình cháu ngu mấy bài chứng minh, so sánh mà mai lại thi Toán rồi.
\(S=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{n.\left(n+3\right)}.\)Với n thuộc N*. Chứng minh S < 1
Ta có : \(S=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{n\left(n+3\right)}\)
\(\Leftrightarrow S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(S=\frac{1}{1}-\frac{1}{n+3}\)
\(S=\frac{n+3}{n+3}-\frac{1}{n+3}=\frac{n+3-1}{n+3}=\frac{n+2}{n+3}<1\)