\(A=4+2^2+2^3+...+2^{2006}\)

\(\mathsf{Đặt}:B=2^2+2^3+...+2^{2006}\\2B=2^3+2^4+...+2^{2007}\\2B-B=(2^3+2^4+...+2^{2007})-(2^2+2^3+...+2^{2006})\\B=2^{2007}-2^2\\B=2^{2007}-4\)

Thay \(B=2^{2007}-4\) vào A, ta được:

\(A=4+(2^{2007}-4)\\\Rightarrow A=2^{2007}\)

$\Rightarrow A$ là 1 luỹ thừa của cơ số 2.

Vậy: ...

1.

a.\(A=1+2^1+2^2+2^3+...+2^{2007}\)

\(2A=2+2^2+2^3+....+2^{2008}\)

b. \(A=\left(2+2^2+2^3+...+2^{2008}\right)-\left(1+2^1+2^2+..+2^{2007}\right)\)

\(=2^{2008}-1\) (bạn xem lại đề)

2.

\(A=1+3+3^1+3^2+...+3^7\)

a. \(2A=2+2.3+2.3^2+...+2.3^7\)

b.\(3A=3+3^2+3^3+...+3^8\)

\(2A=3^8-1\)

\(=>A=\dfrac{2^8-1}{2}\)

3

.\(B=1+3+3^2+..+3^{2006}\)

a. \(3B=3+3^2+3^3+...+3^{2007}\)

b. \(3B-B=2^{2007}-1\)

\(B=\dfrac{2^{2007}-1}{2}\)

4.

Sửa: \(C=1+4+4^2+4^3+4^4+4^5+4^6\)

a.\(4C=4+4^2+4^3+4^4+4^5+4^6+4^7\)

b.\(4C-C=4^7-1\)

\(C=\dfrac{4^7-1}{3}\)

5.

\(S=1+2+2^2+2^3+...+2^{2017}\)

\(2S=2+2^2+2^3+2^4+...+2^{2018}\)

\(S=2^{2018}-1\)

4:

a:Sửa đề: C=1+4+4^2+4^3+4^4+4^5+4^6

=>4*C=4+4^2+...+4^7

b: 4*C=4+4^2+...+4^7

C=1+4+...+4^6

=>3C=4^7-1

=>\(C=\dfrac{4^7-1}{3}\)

5:

2S=2+2^2+2^3+...+2^2018

=>2S-S=2^2018-1

=>S=2^2018-1

\(S=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2005}}\)

\(2.S=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)

\(2.S-S=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\)

\(S=2-\dfrac{1}{2^{2006}}\)

\(S=1-2+2^2-2^3+...+2^{2012}-2^{2013}\)

\(\Rightarrow2S=2-2^2+2^3-2^4+...+2^{2013}-2^{2014}\)

\(\Rightarrow2S+S=2-2^2+2^3-...-2^{2014}+1-2^2-2^3+...-2^{2013}\)

\(\Rightarrow3S=1-2^{2014}\)\(\Rightarrow3S-2^{2014}=1-2^{2015}\)

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

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

\(2S=2+2^2+...+2^{2022}\\ \Leftrightarrow2S-S=S=2^{2022}-1\)