giúp minh

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

\(B=2+2^2+2^3+...+2^{100}\)

\(\Rightarrow2B=2^2+2^3+2^4+...+2^{101}\)

\(\Rightarrow2B-B=2^2+2^3+2^4+...+2^{101}-2-2^2-2^3-...-2^{100}\)

\(\Rightarrow2B-B=2^{101}-2\)

bài 1

2¹⁰¹ - 2

\(A=1+3+3^2+3^3+...+3^{99}\\ \Rightarrow A=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)

\(\Rightarrow A=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{96}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow A=\left(1+3+3^2+3^3\right)\left(1+3^4+...+9^{96}\right)\)

\(\Rightarrow A=40\left(1+3^4+...+9^{96}\right)⋮40\)

Có : \(S=1+2+2^2+2^3+....+2^{99}\)

\(\Rightarrow2S=2+2^2+2^3+....+2^{100}\)

\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+....+2^{99}\right)\)

\(\Rightarrow S=2^{100}-1< 2^{100}\)

Vậy \(S< 2^{100}\)

 S=1+2+2²+2³+....+2⁹⁹

⇒2S=2+2²+2³+....+2¹⁰⁰

⇒2S−S=2¹⁰⁰-1

S=2¹⁰⁰-1

vì 2¹⁰⁰ -1<2¹⁰⁰

⇒S<2¹⁰⁰

a) \(A=1+2+2^2+...+2^{50}\)

\(\Rightarrow2A=2+2^2+...+2^{51}\)

\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)

b) \(B=1+3+3^2+...+3^{100}\)

\(\Rightarrow3B=3+3^2+...+3^{101}\)

\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)

\(\Rightarrow B=\dfrac{3^{101}-1}{2}\)

c) \(C=5+5^2+...+5^{30}\)

\(\Rightarrow5C=5^2+5^3+...+5^{31}\)

\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)

\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)

d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)

\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)

\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)

\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)

a: Tổng các số hạng là:

\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)

Ta có: A+1=2x

\(\Leftrightarrow2x=24311\)

hay \(x=\dfrac{24311}{2}\)

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

a.

$S=1+2+2^2+2^3+...+2^{2017}$
$2S=2+2^2+2^3+2^4+...+2^{2018}$

$\Rightarrow 2S-S=(2+2^2+2^3+2^4+...+2^{2018}) - (1+2+2^2+2^3+...+2^{2017})$

$\Rightarrow S=2^{2018}-1$

b.

$S=3+3^2+3^3+...+3^{2017}$
$3S=3^2+3^3+3^4+...+3^{2018}$

$\Rightarrow 3S-S=(3^2+3^3+3^4+...+3^{2018})-(3+3^2+3^3+...+3^{2017})$

$\Rightarrow 2S=3^{2018}-3$
$\Rightarrow S=\frac{3^{2018}-3}{2}$
 

Câu c, d bạn làm tương tự a,b. 

c. Nhân S với 4. Kết quả: $S=\frac{4^{2018}-4}{3}$

d. Nhân S với 5. Kết quả: $S=\frac{5^{2018}-5}{4}$

C gbcgghfdhsgxwvdgdrgdtdgst

\(A=1+2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(\Rightarrow2A=2+2^2+2^3+2^4+2^5+...+2^{100}+2^{101}\)

\(\Rightarrow2A-A=2^{101}-1\)

\(\Leftrightarrow A=2^{101}-1\)

Đặt biểu thức là A

ta có 2A-A=2^101-1

Đặt \(A=1+2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{100}+2^{101}\)

\(\Rightarrow A=2A-A=\left(2+2^2+2^3+2^4+...+2^{101}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)=2^{101}-1\)