cho S=1/31+1/32+1/33+...+1/60 chung minh 3/5 < S < 4/5
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LH
2
16 tháng 4 2022
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)
Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ; (1/51 + 1/52+...+1/59+1/60) > 1/6
S > 1/4 + 1/5 + 1/6.
Trong khi đó (1/4 + 1/5 + 1/6) > 3/5
=>S > 3/5 (1)
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)
=> S < 4/5 (2)
Từ (1) và (2) => 3/5 <S<4/5 Chúc bạn học tốt !
O
1
15 tháng 6 2023
S=(1/31+1/32+...+1/40)+(1/41+...+1/50)+(1/51+...+1/60)
=>S>1/40*10+1/50*10+1/60*10=3/5
S=(1/31+1/32+...+1/40)+(1/41+...+1/50)+(1/51+...+1/60)
=>S<1/30*10+1/40*10+1/50*10=4/5
=>3/5<S<4/5
NH
0
NX
0
28 tháng 4 2023
1/31>1/40
1/32>1/40
...
1/40=1/40
=>1/31+1/32+...+1/40>1/40*10=1/4
1/41>1/50
1/42>1/50
...
1/50=1/50
=>1/41+1/42+...+1/50>10/50=1/5
1/51>1/60
1/52>1/60
...
1/60=1/60
=>1/51+1/52+...+1/60>10/60=1/6
=>S>1/4+1/5+1/6=3/5
1/31<1/30
1/32<1/30
...
1/40<1/30
=>1/31+1/32+...+1/40<1/30*10=1/3
1/41<1/40
1/42<1/40
...
1/50<1/40
=>1/41+1/42+...+1/50<10/40=1/4
1/51<1/50
1/52<1/50
...
1/60<1/50
=>1/51+1/52+...+1/60<10/50=1/5
=>S<1/3+1/4+1/5=4/5
NM
0
NH
1
AN
0
NM
1
7 tháng 7 2016
A = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)
Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ; (1/51 + 1/52+...+1/59+1/60) > 1/6
A > 1/4 + 1/5 + 1/6.
Trong khi đó (1/4 + 1/5 + 1/6) > 3/5
Vậy A > 3/5 (1)
Mặt khác
A = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)
Tương tự : (1/41 + 1/42 + ...+ 1/50) < 1/4 ; (1/51 + 1/52+...+1/59+1/60) < 1/5
Mà A = (1/3 + 1/4 + 1/5) < 4/5 (Vì 1/3 + 1/5 < 3/5 hay 7/12 < 3/5 hay 35/60 < 36/60)
Vậy A < 4/5 (2)
Từ (1);(2)=> 3/5 <S <4/5 (dpcm)
NM
2
N3
11 tháng 4 2019
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng) Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ; (1/51 + 1/52+...+1/59+1/60) > 1/6 S > 1/4 + 1/5 + 1/6.
Trong khi đó (1/4 + 1/5 + 1/6) > 3/5 =>S > 3/5 (1) S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60) Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng) => S < 4/5 (2)
Từ (1) và (2) => 3/5 <S<4/5
BA
11 tháng 4 2019
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+...+\frac{1}{60}\right)\)
Mà \(\left(\frac{1}{31}+...+\frac{1}{40}\right)>\frac{1}{40}\cdot10=\frac{1}{4}\)
Tương tự : \(\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)>\frac{1}{5}\)
\(\left(\frac{1}{51}+...+\frac{1}{60}\right)>\frac{1}{6}\)
\(S>\frac{1}{4}+\frac{1}{5}+\frac{1}{6}>\frac{3}{5}\)(*1)
Mặt khác:\(\left(\frac{1}{31}+...+\frac{1}{40}\right)< \frac{1}{31}\cdot10=\frac{1}{3}\)
\(\Rightarrow S< \frac{4}{5}\)(*2)
Từ (*1)(*2)= 3/5<S<4/5
\(S=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\) (có 30 số hạng)
\(\Rightarrow S=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)(có 10 số hạng mỗi nhóm)
Ta có:
+) \(\left\{{}\begin{matrix}\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}>\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{10}{40}=\dfrac{1}{4}\\\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}>\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{10}{50}=\dfrac{1}{5}\\\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{10}{60}=\dfrac{1}{6}\end{matrix}\right.\)
\(\Rightarrow S=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)< \dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{37}{60}\)
Mà \(\dfrac{37}{60}>\dfrac{36}{60}=\dfrac{3}{5}\) nên:
\(S>\dfrac{3}{5}\) (1)
+) \(\left\{{}\begin{matrix}\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}< \dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}=\dfrac{10}{30}=\dfrac{1}{3}\\\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{10}{40}=\dfrac{1}{4}\\\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{10}{50}=\dfrac{1}{5}\end{matrix}\right.\)
\(\Rightarrow S=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)< \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}\)
Mà \(\dfrac{47}{60}< \dfrac{48}{60}=\dfrac{4}{5}\) nên:
\(S< \dfrac{4}{5}\) (2)
Từ (1) và (2) suy ra:
\(\dfrac{3}{5}< S< \dfrac{4}{5}\)
Vậy...