A = 1 + 2 + 2² + 2³ + ... + 2⁶² + 2⁶³

2A = 2 + 2² + 2³ + 2⁴ + ... + 2⁶³ + 2⁶⁴

A = 2⁶⁴ - 1

\(S_1=1+2+2^2+2^3+..+2^{63}\\ \Rightarrow2S_1=2+2^2+2^3+2^4+...+2^{64}\\ \Rightarrow S_1-2S_1=1-2^{64}\\ \Rightarrow-S_1=1-2^{64}\\ \Rightarrow S_1=2^{64}-1.\)

- Ta có: S₁ = 1 + 2 + 2² + 2³ + … + 2⁶³ = 1 + 2(1 + 2 + 2² + 2³ + … + 2⁶²) (1)

= 1 + 2(S₁ - 2⁶³) = 1 + 2S₁ - 2⁶⁴ S₁ = 2⁶⁴ - 1

H2.right

`#3107.101107`

`S_1 = 1 + 2 + 2^2 + 2^3 + ... + 2^63`

`2S_1 = 2 + 2^2 + 2^3 + .... + 2^64`

`2S_1 - S_1 = (2 + 2^2 + 2^3 + ... + 2^64) - (1 + 2 + 2^2 + 2^3 + ... + 2^63)`

`S_1 = 2 + 2^2 + 2^3 + ... + 2^64 - 1 - 2 - 2^2 - 2^3 - ... - 2^63`

`S_1 = 2^64 - 1`

Vậy, `S_1 = 2^64 - 1.`

tính như nào nhỉ

`1+2+2^2+2^3+....+2^63`

`=2+2+2^2+2^3+....+2^63-1`

`=2.2+2^2+2^3+....+2^63-1`

`=2^2+2^2+2^3+....+2^63-1`

`=2.2^2+2^3+....+2^63-1`

`=2^3+2^3+...2^63-1`

`=2.2^3+....+2^63-1`

`=2^4+....+2^63-1`

`=2^{63}.2-1=2^64-1`

Đặt \(A=1+2+2^2+2^3+...+2^{63}\)

\(\Rightarrow2A=2+2^2+2^3+2^4+2^5+...+2^{64}\)

\(\Rightarrow A=2A-A=\left(2+2^2+2^3+2^4+...+2^{64}\right)-\left(1+2+2^2+2^3+...+2^{63}\right)=2^{64}-1\left(đpcm\right)\)

\(=526+1590-262+795=2649\)

Lời giải:

$A=(1+2)+(2^2+2^3)+....+(2^{2020}+2^{2021})$

$=3+2^2(1+2)+....+2^{2020}(1+2)$

$=3+3.2^2+....+3.2^{2020}$

$=3(1+2^2+....+2^{2020})\vdots 3$
Ta có đpcm.

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.

\(H=3+3^2+...+3^{2022}\)

\(\Rightarrow3H=3^2+3^3+...+3^{2023}\)

\(\Rightarrow3H-H=3^{2023}-3\)

\(\Rightarrow2H=3^{2023}-3\)

\(\Rightarrow H=\dfrac{3^{2023}-3}{2}\)

Bài 2

H = 3 + 3² + 3³ + ... + 3²⁰²²

⇒ 3H = 3² + 3³ + 3⁴ + ... + 3²⁰²³

⇒2H = 3H - H

= (3² + 3³ + 3⁴ + ... + 3²⁰²³) - (3 + 3² + 3³ + ... + 3²⁰²²)

= 3²⁰²³ - 3

⇒ H = (3²⁰²³ - 3) : 2

a) \(S=1+2+2^2+2^3+...+2^{2022}=\dfrac{2^{2022+1}-1}{2-1}=2^{2023}-1\)

b) \(S=1+4+4^2+4^3+...+4^{2022}=\dfrac{4^{2022+1}-1}{4-1}=\dfrac{4^{2023}-1}{3}\)

\(S=1+2+2^2+2^3+...+2^{2022}\\ 2S=2+2^2+2^3+2^4+...+2^{2023}\\ 2S-S=2+2^2+2^3+2^4+...+2^{2023}-1-2-2^2-2^3-...-2^{2022}\\ S=2^{2023}-1\\ S=4+4^2+4^3+...+4^{2022}\\ 4S=4^2+4^3+4^4+...+4^{2023}\\ 4S-S=4^2+4^3+4^4+...+4^{2023}-4-4^2-4^3-...-4^{2023}\\ 3S=4^{2023}-4\\ S=\dfrac{4^{2023}-4}{3}\)

\(A=\left(1+2\right)+2^2\left(1+2\right)+...+2^{10}\left(1+2\right)=3+2^2.3+...+2^{10}.3=3\left(1+2^2+...+2^{10}\right)⋮3\)

\(A=1+2+2^2+2^3+...+2^{10}+2^{11}\)

\(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{10}\left(1+2\right)\)

\(=\left(1+2\right)\left(1+2^2+...+2^{10}\right)\)

\(=3\left(1+2^2+...+2^{10}\right)\) ⋮3

`#3107`

\(A=1+2^1+2^2+2^3+...+2^{2015}\)

\(2A=2+2^2+2^3+2^4+...+2^{2016}\)

\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2016}\right)-\left(1+2+2^2+2^3+...+2^{2015}\right)\)

\(A=2+2^2+2^3+2^4+...+2^{2016}-1-2-2^2-2^3-...-2^{2015}\)

\(A=2^{2016}-1\)

Vậy, \(A=2^{2016}-1.\)

\(A=2^0+2^1+2^2+...+2^{2015}\)

\(2\cdot A=2^1+2^2+2^3+...+2^{2016}\)

\(A=2A-A=2^{2016}-2^0\)

\(A=2^{2016}-1\)