Cho S = 1 2 + 1 6 + 1 12 + 1 20 + 1 30 + 1 42 . So sánh S với 1.
S= 1/2 + 1/6 + 1/12 + 1/20 +... + 1/2352 + 1/2450 (phân số)
=(100+121+144):(169+196)
=365:365
=1
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2352}+\frac{1}{2450}\\ \)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
S= 1 phần 2 cộng 1 phần 6 cộng 1 phần 12 cộng 1 phần 20 ... 1 phần 380
`S=1/2 +1/6 +1/12 +1/20 +...+1/380`
`=1/(1.2)+1/(2.3) +1/(3.4)+1/(4.5)+...+1/(19.20)`
`=1-1/2 +1/2 -1/3 +1/3-1/4 +1/4 -1/5+....+1/19-1/20`
`=1-1/20=20/20 -1/20 =19/20`
cho s=1/11+1/12+...+1/20 chứng minh 1/2<s<1
Tính tổng: S=2/3.7+2/7.1+2/11.15+...+2/95.99
M=1/2+1/6+1/12+1/20+1/110
S=2/3.7+2/7.1+2/11.15+...+2/95.99
S = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{11}\right)+\frac{1}{2}.\left(\frac{1}{11}-\frac{1}{15}\right)+...+\frac{1}{2}.\left(\frac{1}{95}-\frac{1}{99}\right)\)
S = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\right)\)
S = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
S = \(\frac{1}{2}.\frac{32}{99}\)
S = \(\frac{16}{99}\)
M=1/2+1/6+1/12+1/20+..+1/110
M = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\)
M = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
M = \(1-\frac{1}{11}\)
M = \(\frac{10}{11}\)
tính tổng S=1/6+1/12+1/20+..........+1/600+1/650
\(S=\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{600}+\dfrac{1}{650}\)
\(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{24\cdot25}+\dfrac{1}{25\cdot26}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}\)
\(=\dfrac{1}{2}-\dfrac{1}{26}=\dfrac{13-1}{26}=\dfrac{12}{26}=\dfrac{6}{13}\)
s= 1/11+1/12+1/13+...+1/20 chứng minh s<5/6
#)Giải :
Ta có : \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}< \frac{5}{6}\)(có 10 số \(\frac{1}{20}\))
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}< \frac{5}{6}\)
Hay \(S< \frac{5}{6}\left(đpcm\right)\)
S=1/11+1/12+1/13+1/14+...+1/20 so sanh S voi 5/6
Ta có :S=1/11+1/12+1/13+...+1/20<1/10+1/10+1/10+...+1/10(20 số hạng 1/10)
=>S<1/10.20=1/2<5/6
ĐPCM
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
Ta có \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}\)
\(=\frac{5}{6}\)
Cho S= 1/11 + 1/12 + 1/13 +.....+ 1/20
CMR:1/2 < S <1
Ta thấy: \(\frac{1}{11}>\frac{1}{20}\)
\(\frac{1}{12}>\frac{1}{20}\)
\(\frac{1}{13}>\frac{1}{20}\)
................
\(\frac{1}{19}>\frac{1}{20}\)
=>\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}+\frac{1}{20}\)
=>\(S>\frac{10}{20}\)
=>\(S>\frac{1}{2}\)(1)
Lại có:
\(\frac{1}{11}<\frac{1}{10}\)
\(\frac{1}{12}<\frac{1}{10}\)
\(\frac{1}{13}<\frac{1}{10}\)
................
\(\frac{1}{20}<\frac{1}{10}\)
=>\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}<\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)
=>\(S<\frac{10}{10}\)
=>S<1 (2)
Từ (1) và (2)
=>1/2<S<1
=>ĐPCM
Ta có:
1/20=1/20
1/19>1/20
1/18>1/20
.....
1/11>1/20
Suy ra 1/11+1/12+1/13+...+1/20>1/20*10=1/2. Vậy S>1/2.
Tương tự, ta có:
1/11=1/11
1/12<1/11
1/13<1/11
....
1/20<1/11
Suy ra 1/11+1/12+1/13+...+1/20<1/11*10=10/11<11/11=1. Vậy S < 1.
VẬY 1/2< S<1
Cho S= 1/11 + 1/12 + 1/13 +.....+ 1/20
CMR:1/2 < S <1