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Bài 1
a) S = 1 + 2 + 2² + 2³ + ... + 2²⁰²³
2S = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²⁴
S = 2S - S = (2 + 2² + 2³ + ... + 2²⁰²⁴) - (1 + 2 + 2² + 2³)
= 2²⁰²⁴ - 1
b) B = 2²⁰²⁴
B - 1 = 2²⁰²⁴ - 1 = S
B = S + 1
Vậy B > S
a,
\(S=1+2+2^2+...+2^{2023}\)
\(2S=2+2^2+2^3+...+2^{2024}\)
\(\Rightarrow S=2^{2024}-1\)
b.
Do \(2^{2024}-1< 2^{2024}\)
\(\Rightarrow S< B\)
2.
\(H=3+3^2+...+3^{2022}\)
\(\Rightarrow3H=3^2+3^3+...+3^{2023}\)
\(\Rightarrow3H-H=3^{2023}-3\)
\(\Rightarrow2H=3^{2023}-3\)
\(\Rightarrow H=\dfrac{3^{2023}-3}{2}\)
Giải:
A=1/22+1/32+1/42+...+1/92
Ta có:
1/22<1/1.2
1/32<1/2.3
1/42<1/3.4
...
1/92<1/8.9
⇒A<1/1.2+1/2.3+1/3.4+...+1/8.9
A<1/1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9
A<1/1-1/9
A<8/9
Ta có:
1/22>1/2.3
1/32>1/3.4
1/42>1/4.5
...
1/92>1/9.10
⇒A>1/2.3+1/3.4+1/4.5+...+1/9.10
A>1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10
A>1/2-1/10
A>2/5
Vậy 2/5<A<8/9 (đpcm)
Chúc bạn học tốt!
\(32^{15}=\left(2^5\right)^{15}=2^{5.15}=2^{75}\)
\(4^{39}=\left(2^2\right)^{39}=2^{2.39}=2^{78}\)
Do \(2^{78}>2^{75}\)
\(\Rightarrow4^{39}>32^{15}\)
\(\Rightarrow1+4+4^2+...+4^{39}>32^{15}\)
\(\Rightarrow3\left(1+4+4^2+...+4^{39}\right)>32^{15}\)
Vậy \(A>B\)
\(A=\frac{1}{31}+\frac{1}{33}+\frac{1}{35}+\frac{1}{63}+\frac{1}{65}+\frac{1}{67}+\frac{1}{91}+\frac{1}{92}+\frac{1}{94}\)
\(< \frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\)
\(=\frac{9}{30}=\frac{3}{10}< \frac{4}{10}=\frac{2}{5}\).
a) Ta có: 222 = 219.23 = 8. 219 > 7. 219
=> 7.2^19 < 2^22
b)
Ta có: \(32^{15}=\left(2^5\right)^{15}=2^{75}\)
\(8^{25}=\left(2^3\right)^{25}=2^{75}\)
=> 32^15 và 8^25
A = 1 + 3 + 3² + ... + 3²⁰²³
⇒ 3A = 3 + 3² + 3³ + ... + 3²⁰²³ + 3²⁰²⁴
⇒ 2A = 3A - A
= (3 + 3² + 3³ + ... + 3²⁰²³ + 3²⁰²⁴) - (1 + 3 + 3² + ... + 3²⁰²³)
= 3²⁰²⁴ - 1
⇒ A = (3²⁰²⁴ - 1) : 2
⇒ A < B
A=1+3+32+33+34+........+32022+32023
3A=3+32+33+............+32023+32024
3A-A=(3+32+33+..........+32023+32024
A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰
⇒ 2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹
⇒ A = 2A - A = (2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹) - (2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰)
= 2²⁰¹¹ - 2⁰
= 2²⁰¹¹ - 1
= B
Vậy A = B
Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)và B= \(\frac{8}{9}\)
Ta có :\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)
\(........\)
\(\frac{1}{9^2}< \frac{1}{8\cdot9}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{9^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{8\cdot9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{9^2}=\frac{8}{9}\)
Vậy \(A=B\)