Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: Ta có
A = \(\dfrac{1}{10}\) + \((\dfrac{1}{11}\) + \(\dfrac{1}{12}\) + ...+ \(\dfrac{1}{100}\)\()\)
⇒ A > \(\dfrac{1}{10}\) + \((\dfrac{1}{100}\) + \(\dfrac{1}{100}\) + ...+ \(\dfrac{1}{100}\)\()\)90 số hạng
⇒ A > \(\dfrac{1}{10}\) + \(\dfrac{90}{100}\)
⇒ A > 1
vậy A > 1
b: ta có
S = (\(\dfrac{1}{21}\) + \(\dfrac{1}{22}\)+ \(\dfrac{1}{23}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{26}\) + \(\dfrac{1}{27}\)+ \(\dfrac{1}{28}\) + \(\dfrac{1}{29}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{33}\) + \(\dfrac{1}{34}\) + \(\dfrac{1}{35}\))
⇒ S > (\(\dfrac{1}{25}\) + \(\dfrac{1}{25}\)+ \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{30}\) + \(\dfrac{1}{30}\)+ \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{35}\) + \(\dfrac{1}{35}\)+ \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\))
⇔ S > \(\dfrac{5}{25}\)+\(\dfrac{5}{30}\)+\(\dfrac{5}{35}\)
⇔ S > \(\dfrac{1}{5}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{7}\)
⇔ S > \(\dfrac{107}{210}\)> \(\dfrac{105}{210}\)=\(\dfrac{1}{2}\)
vậy S > \(\dfrac{1}{2}\)
Ta có :
Cần 30 số hạng đầu đã lớn hơn 1.
1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2
1/20+1/21+…+1/29 > 1/30+1/30+…+1/30 = 10/30 = 1/3
1/30+1/31+…+1/39 > 1/40+1/40+…+1/40 = 10/40 = 1/4
=> 1/10+1/11+…+1/39 > 1/2+1/3+1/4 = 13/12 > 1
Vậy :C>1
1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2
1/20+1/21+…+1/29 > 1/30+1/30+…+1/30 = 10/30 = 1/3
1/30+1/31+…+1/39 > 1/40+1/40+…+1/40 = 10/40 = 1/4
=> A>1
Chỉ cần 30 số hạng đầu đã lớn hơn 1.
1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2
1/20+1/21+…+1/29 > 1/30+1/30+…+1/30 = 10/30 = 1/3
1/30+1/31+…+1/39 > 1/40+1/40+…+1/40 = 10/40 = 1/4
=>
1/10+1/11+…+1/39 > 1/2+1/3+1/4 = 13/12 > 1
A = 1 / 10 + ( 1 / 11 + 1 / 12 + ... + 1 / 99 + 1 / 100 )
A = 1 / 10 + ( 1 / 11 + 1 / 12 + ... + 1 / 99 + 1 / 100 ) > 1 / 10 + ( 1 / 100 + 1 / 100 + ... + 1 / 100 )
= 1 / 10 + 90 / 100 = 1
Vậy A > 1
1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2
1/20+1/21+…+1/29 > 1/30+1/30+…+1/30 = 10/30 = 1/3
1/30+1/31+…+1/39 > 1/40+1/40+…+1/40 = 10/40 = 1/4
=>
1/10+1/11+…+1/39 > 1/2+1/3+1/4 = 13/12 > 1
đúng nhé
\(C=\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
\(>\frac{1}{50}.41+\frac{1}{100}.50=\frac{41}{50}+\frac{50}{100}=\frac{33}{25}=1\frac{8}{25}>1\)
Ta thấy rằng mỗi số hạng trong tổng đều lớn hơn hoặc bằng \(\frac{1}{100}\)
=> \(C>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{1}{100}x100=1\)
=> C>1 (Đpcm)
a) P = 1 + 3 + 3² + ... + 3¹⁰¹
= (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3⁹⁹ + 3¹⁰⁰ + 3¹⁰¹)
= 13 + 3³.(1 + 3 + 3²) + ... + 3⁹⁹.(1 + 3 + 3²)
= 13 + 3³.13 + ... + 3⁹⁹.13
= 13.(1 + 3³ + ... + 3⁹⁹) ⋮ 13
Vậy P ⋮ 13
b) B = 1 + 2² + 2⁴ + ... + 2²⁰²⁰
= (1 + 2² + 2⁴) + (2⁶ + 2⁸ + 2¹⁰) + ... + (2²⁰¹⁶ + 2²⁰¹⁸ + 2²⁰²⁰)
= 21 + 2⁶.(1 + 2² + 2⁴) + ... + 2²⁰¹⁶.(1 + 2² + 2⁴)
= 21 + 2⁶.21 + ... + 2²⁰¹⁶.21
= 21.(1 + 2⁶ + ... + 2²⁰¹⁶) ⋮ 21
Vậy B ⋮ 21
c) A = 2 + 2² + 2³ + ... + 2²⁰
= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2¹⁷ + 2¹⁸ + 2¹⁹ + 2²⁰)
= 30 + 2⁴.(2 + 2² + 2³ + 2⁴) + ... + 2¹⁶.(2 + 2² + 2³ + 2⁴)
= 30 + 2⁴.30 + ... + 2¹⁶.30
= 30.(1 + 2⁴ + ... + 2¹⁶)
= 5.6.(1 + 2⁴ + ... + 2¹⁶) ⋮ 5
Vậy A ⋮ 5
d) A = 1 + 4 + 4² + ... + 4⁹⁸
= (1 + 4 + 4²) + (4³ + 4⁴ + 4⁵) + ... + (4⁹⁷ + 4⁹⁸ + 4⁹⁹)
= 21 + 4³.(1 + 4 + 4²) + ... + 4⁹⁷.(1 + 4 + 4²)
= 21 + 4³.21 + ... + 4⁹⁷.21
= 21.(1 + 4³ + ... + 4⁹⁷) ⋮ 21
Vậy A ⋮ 21
e) A = 11⁹ + 11⁸ + 11⁷ + ... + 11 + 1
= (11⁹ + 11⁸ + 11⁷ + 11⁶ + 11⁵) + (11⁴ + 11³ + 11² + 11 + 1)
= 11⁵.(11⁴ + 11³ + 11² + 11 + 1) + 16105
= 11⁵.16105 + 16105
= 16105.(11⁵ + 1)
= 5.3221.(11⁵ + 1) ⋮ 5
Vậy A ⋮ 5
