Ta có :

B = 2¹⁰⁰ - 2⁹⁹ + 2⁹⁸ - 2⁹⁷ + ... + 2² - 2 + 1

=> B = ( 2¹⁰⁰ + 2⁹⁸ + ... + 2² + 1 ) - ( 2⁹⁹ + 2⁹⁷ + ... + 2 )

=> 2²B = 2 . [ ( 2¹⁰⁰ + 2⁹⁸ + ... + 2² + 1 ) - ( 2⁹⁹ + 2⁹⁷ + ... + 2 ) ]

=> 4B = ( 2¹⁰² + 2¹⁰⁰ + ... + 2² ) - ( 2¹⁰¹ + 2⁹⁹ + ... + 2³ )

=> 4B - B = [( 2¹⁰² + 2¹⁰⁰ + ... + 2² ) - ( 2¹⁰¹ + 2⁹⁹ + ... + 2³ )] - [( 2¹⁰⁰ + 2⁹⁸ + ... + 2² + 1 ) - ( 2⁹⁹ + 2⁹⁷ + ... + 2 )]

=> 3B = ( 2¹⁰² - 1 ) + ( 2 - 2¹⁰¹ )

=> 3B = 2¹⁰¹ - 1

=> B = \(\frac{2^{101} - 1}{3}\)

gọi dãy số là A, ta có:

A = 2^{100 -} 2^{99 - ...... -} 2¹

2A = 2^{101 -} 2^{100 - .... -} 2²

2A = ( 2¹⁰¹ - ... - 2^{2 ) -} ( 2¹⁰⁰ - ... - 2 )

A = 2^{101 -} 2

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

\(3^2S=9S=3^2+3^4+3^6+...+3^{2024}\)

\(S=\dfrac{9S-S}{8}=\left(3^{2024}-1\right):8\)

d, không đáp án nào đúng

Lời giải:

$S=1+3^2+3^4+....+3^{2022}$

$9S=3^2S=3^2+3^4+3^6+...+3^{2024}$

$\Rightarrow 9S-S=3^{2024}-1$

$\Rightarrow S=\frac{3^{2024}-1}{8}$

Đáp án D.

a) \(A=2+2^2+2^3+...+2^{2017}\)

\(A=2\left(1+2^1+2^2+...+2^{2016}\right)\)

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

\(A=2.\left(2^{2017}-1\right)=2^{2018}-2\)

Câu b bạn xem lại đề

a)

1/ 3^x-1 + 5.3^x-1 = 162

3^x-1(1 + 5) = 162

3^x-1 = \(\frac{162}{6}\)

3^x-1 = 27

3^x-1 = 3³

x - 1 = 3

x = 4

2/ B = 3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1

\(\Rightarrow\) 3B = 3.3¹⁰⁰ - 3.3⁹⁹ + 3.3⁹⁸ - 3.3⁹⁷ + ... + 3.3² - 3.3 + 3.1

= 3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸ + ... + 3³ - 3² + 3

Ta có:

4B = 3B + B = (3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸ + ... + 3³ - 3² + 3) + (3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1)

= 3¹⁰¹ + 3¹⁰⁰ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁹ + 3⁹⁸ - 3⁹⁸ + ... + 3 - 3 + 1

= 3¹⁰¹ + 1

\(\Rightarrow\) B = \(\frac{3^{101}+1}{4}\)

Mình làm ngắn gọn nhé.

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

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

\(\Rightarrow2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}\)

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

\(B=1+3+...+3^{66}\)

\(3B=3+3^2+...+3^{67}\)

\(2B=3+3^2+...+3^{67}-1-3-...-3^{66}\)

\(2B=3^{67}-1\)

\(B=\frac{3^{67}-1}{2}\)

Lời giải :

1. \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)

\(=\frac{a^3}{8}+\frac{3a^2b}{4}+\frac{3ab^2}{2}+b^3+\frac{a^3}{8}-\frac{3a^2b}{4}+\frac{3ab^2}{2}-b^3\)

\(=\frac{a^3}{4}+3ab^2\)

Lời giải :

2. \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy...

3. \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=64x^3-48x^2+12x-1-64x^3+48x^2-12x+9\)

\(=8\)