Cho S = 1/5+1/6+1/7+......+1/19 . So sánh S với 2
So sánh S với 2 biết :
S=1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8
S=1/2+1/3+1/4+1/5+1/6+1/7+1/8=
=481/280=1,717(857142)
=> S<2
Bài 7. Cho S = 1 + 2 + 2^2 + 2^3 + ..... 2^20.So sánh với 5.2^19
S=1+2+2^2+2^3+...+2^20
2.S=2+2^2+2^3+...+2^20+2^21
2.S-S=S=(2+2^2+2^3+....+2^21)-(1+2+2^2+...+2^20)
S=2^21-1
bây giờ so sánh 2^21-1 với 5.2^19
mà 2^21-1=2^19.2^2-1 hay 2^19 .4 -1 <2^19.5
=>S<2^19.5
\(S=1+2+2^2+2^3+\cdots+2^{20}\)
=> \(2S=2+2^2+2^3+2^4+\cdots+2^{21}\)
=> \(2S-S=\left(2+2^2+2^3+2^4+\cdots+2^{21}\right)-\left(1+2+2^2+2^3+\ldots+2^{20}\right)\)
=>\(S=2^{21}-1\)
Mà \(2^{21}-1=2^{19}.2^2-1\) hay \(2^{19}.4-1<2^{19}.5\)
=>\(S<2^{19}.5\)
Ta có: \(S=1+2+2^2+\cdots+2^{20}\)
=>\(2S=2+2^2+2^3+\cdots+2^{21}\)
=>\(2S-S=2+2^2+\cdots+2^{21}-1-2-\cdots-2^{20}\)
=>\(S=2^{21}-1\)
\(2^{21}-1-5\cdot2^{19}=2^{19}\left(2^2-5\right)-1=-2^{19}-1<0\)
=>\(S-5\cdot2^{19}<0\)
=>\(S<5\cdot2^{19}\)
S=1/6^0 + 1/6^1+1/6^2 +...+1/6^2016+1/6^2017
So sánh S với 6/7
Cho S = 1/11 +1/12+1/13+1/14+1/15+1/16+1/17+1/18+1/19+1/20. So sánh S với 1/2
Ta có:\(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};\frac{1}{13}>\frac{1}{20};....;\frac{1}{19}>\frac{1}{20}\)
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)(Có 10 phân số \(\frac{1}{20}\))
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{10}{20}\)\(\Leftrightarrow S>\frac{10}{20}\)
Mà \(\frac{10}{20}=\frac{1}{2}\)nên
\(\Rightarrow S>\frac{1}{2}\)
S=1/2+1/12+1/13+1/14+...+1/19+1/20. hãy so sánh S với 1/2 ?
A=(1/2+1/12+1/13+1/14+1/15)+(1/16+1/17+1/18+1/19+1/20)
Thay các phân số trong mỗi nhóm bởi phân số nhỏ nhất, ta có:
A> 1/15.5+1/20.5=1/3+1/4= 7/12>1/2
Suy ra A>1/2
Vậy A> 1/2
cho s = 1/50 + 1/51 + 1/52 + 1/53 + .......... + 1/99 + 1/100 . hãy so sánh s với 5/6 cứu mình với
Sửa đề: \(S=\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\)
Ta có: \(\frac{1}{51}<\frac{1}{50};\frac{1}{52}<\frac{1}{50};\ldots;\frac{1}{75}<\frac{1}{50}\)
Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{25}{50}=\frac12\) (1)
Ta có: \(\frac{1}{76}<\frac{1}{75};\frac{1}{77}<\frac{1}{75};\ldots;\frac{1}{100}<\frac{1}{75}\)
Do đó: \(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}<\frac{1}{75}+\frac{1}{75}+\cdots+\frac{1}{75}=\frac{25}{75}=\frac13\) (2)
Từ (1),(2) suy ra \(\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}\right)<\frac12+\frac13\)
=>\(S<\frac56\)
Cho S=1/2 . 3/4 . 5/6.....9999/10000
So sánh S với 0.01
Vì 1/2<2/3;3/4<4/5;5/6<6/7;....;9999/10000<10000/10001
-->S<2/3.4/5.5/6.....10000/10001
Gọi 2/3.4/5.5/6.....10000/10001 là D và D>S
Có D=2/3.4/5.5/6.....10000/10001
-->S.D=(1/2 . 3/4 . 5/6.....9999/10000).(2/3.4/5.5/6.....10000/10001)
S.D=1/2.2/3.3/4.4/5.5/6.6.7.....9999/10000.10000/10001
S.D=1/10001
Vì S<D nên S.S<S.D hay S.S<1/10001
-->S<1/10001
mà 1/10001<1/100=0.01
-->S<1/100=0.01
-->S<0.01
Vậy
Có \(\frac{1}{2}< \frac{2}{3};\frac{2}{3}< \frac{3}{4};.....;\frac{9999}{10000}< \frac{10000}{10001}\)
\(\Rightarrow S^2< \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\)
\(\Rightarrow S^2< \frac{1}{10001}< \frac{1}{10000}=\left(\frac{1}{100}\right)^2\)
\(\Rightarrow S< \frac{1}{1000}< 0,01\)
Vậy S<0,01
So sánh các phân số sau với 1 : 6 / 8 ; 1/4 ; 5/2 ; 7/3 ; 19 / 19
\(\frac{6}{8}< 1\)vì \(\frac{6}{8}< \frac{8}{8}\)
\(\frac{1}{4}< 1\)vì \(\frac{1}{4}< \frac{4}{4}\)
\(\frac{5}{2}>1\)vì \(\frac{5}{2}>\frac{2}{2}\)
\(\frac{7}{3}>1\)vì \(\frac{7}{3}>\frac{3}{3}\)
\(\frac{19}{19}=1\)vì \(\frac{19}{19}=\frac{19}{19}\)
So sánh A với 2, biết A:
1/5+1/6+1/7+...+1/18+1/19