chứng tỏ rằng : 1/22 + 1/32 + 1/42 + ...+1/1002 < 99/100
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PT
1
D
1
L
28 tháng 4 2022
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
KV
22 tháng 10 2023
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
NH
0
27 tháng 4 2021
a)\(\dfrac{1}{2^2}<\dfrac{1}{1.2}\)
\(\dfrac{1}{3^3}<\dfrac{1}{2.3}\)
\(...\)
\(\dfrac{1}{8^2}<\dfrac{1}{7.8}\)
Vậy ta có biểu thức:
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B<1-\dfrac{1}{8}=\dfrac{7}{8}<1\)
Vậy B < 1 (đpcm)
27 tháng 4 2021
Giải:
a) Ta có:
1/22=1/2.2 < 1/1.2
1/32=1/3.3 < 1/2.3
1/42=1/4.4 < 1/3.4
1/52=1/5.5 < 1/4.5
1/62=1/6.6 < 1/5.6
1/72=1/7.7 < 1/6.7
1/82=1/8.8 <1/7.8
⇒B<1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8
B<1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8
B<1/1-1/8
B<7/8
mà 7/8<1
⇒B<7/8<1
⇒B<1
b)S=3/1.4+3/4.7+3/7.10+...+3/40.43+3/43.46
S=1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46
S=1/1-1/46
S=45/46
Vì 45/46<1 nên S<1
Vậy S<1
Chúc bạn học tốt!
Z
26 tháng 7 2019
Lời giải:
a, Ta có: \(A=\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+...+\frac{1}{22}>\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+...+\frac{1}{22}=\frac{1}{22}.11=\frac{11}{22}=\frac{1}{2}\)
Vậy: \(A>\frac{1}{2}\)
b, Ta có: \(B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{99}+\frac{1}{100}\)
\(=\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Mà: \(\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\text{}\text{}\text{}>\left(\frac{1}{50}+...+\frac{1}{50}+\frac{1}{50}\right)+\left(\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\right)\)
=> \(B\text{}\text{}\text{}>\frac{1}{50}.41+\frac{1}{100}.50=\frac{41+25}{50}=\frac{33}{25}>1\)
Vậy: \(B>1\)
c, Ta có: \(C=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{16}+\frac{1}{17}< \frac{1}{5}+\frac{1}{6}+\left(\frac{1}{7}+...+\frac{1}{7}+\frac{1}{7}\right)=\frac{11}{30}+11.\frac{1}{7}=\frac{407}{210}< \frac{420}{210}=2\)
Vậy: \(C< 2\)
Chúc bạn học tốt!
Tick cho mình nhé!
H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
Ta có:\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
Mà \(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{99}{100}\)