P=1/52+1/62+....+1/1002
chung minh rang:1/6<P<14
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S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)
(*)Ta có:
1/13<1/12
1/14<1/12
1/15<1/12
=>1/13+1/14+1/15<1/12
(*)Ta lại có:
1/61<1/60
1/62<1/60
1/63<1/60
=>1/61+1/62+1/63<1/60
=>S<1/5+1/12.3+1/60.3
S<1/5+1/4+1/20
S<1/2
S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)
(*)Ta có:
1/13<1/12
1/14<1/12
1/15<1/12
=>1/13+1/14+1/15<1/12
(*)Ta lại có:
1/61<1/60
1/62<1/60
1/63<1/60
=>1/61+1/62+1/63<1/60
=>S<1/5+1/12.3+1/60.3
S<1/5+1/4+1/20
S<1/2
Ta có: \(\dfrac{1}{5^2}>\dfrac{1}{5.6};\dfrac{1}{6^2}>\dfrac{1}{6.7};...;\dfrac{1}{100^2}>\dfrac{1}{100.101}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5}-\dfrac{1}{101}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{96}{505}>\dfrac{1}{6}\) (1)
Ta có: \(\dfrac{1}{5^2}< \dfrac{1}{4.5};\dfrac{1}{6^2}< \dfrac{1}{5.6};\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\) (2)
Từ (1) và (2)⇒\(\dfrac{1}{6}< B< \dfrac{1}{4}\)
a: 5A=5+5^2+...+5^2023
=>4A=5^2023-1
=>A=(5^2023-1)/4
b: 6B=6^2+6^3+...+6^41
=>5B=6^41-6
=>B=(6^41-6)/5
c: 16C=4^4+4^6+...+4^16
=>15C=4^16-4^2
=>C=(4^16-4^2)/15
d: 9D=3^3+3^5+...+3^27
=>8D=3^27-3
=>D=(3^27-3)/8
Đặt A=11⋅2+12⋅3+...+17⋅8A=11⋅2+12⋅3+...+17⋅8
Dễ thấy: B=122+132+...+182B=122+132+...+182<A=11⋅2+12⋅3+...+17⋅8(1)<A=11⋅2+12⋅3+...+17⋅8(1)
Ta có:A=11⋅2+12⋅3+...+17⋅8A=11⋅2+12⋅3+...+17⋅8
=1−12+12−13+...+17−18=1−12+12−13+...+17−18
=1−18<1(2)=1−18<1(2)
Từ (1);(2)(1);(2) ta có: B<A<1⇒B<1
\(P=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
- Có: \(P>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)
=> \(P>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
=> \(P>\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)
=> \(P>\frac{1}{6}\)(1)
- Có: \(P< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
=> \(P< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}\)
=> \(P< \frac{1}{4}-\frac{1}{100}< 14\)(2)
Từ (1) và (2)
=> \(\frac{1}{6}< P< 14\)(Nếu đề là 1/6 < P < 1/4 thì thay số 14 bằng 1/4 vẫn đúng nhé)
=> Đpcm