Bài 1:

A = 1 + 3 + 3² + ... + 3¹⁰⁰

=> 3A = 3 + 3² + ... + 3¹⁰¹

=> 2A = 3¹⁰¹ - 1

=> A = \(\frac{3^{101}-1}{2}\)

B = 1 + 4² + 4⁴ + ... + 4¹⁰⁰

=> 8B = 4² + 4⁴ + ... + 4¹⁰²

=> 7B = 4¹⁰² - 1

=> B = \(\frac{4^{102}-1}{7}\)

Bài 2:

a) S₁ = 2² + 4² + ... + 20²

=> S₁ = 2²(1+2²+...+10²)

=> S₁ = 2².385

=> S₁ = 1540

b) S₂ = 100² + 200² + ... + 1000²

=> S₂ = 100²(1+2²+...+10²)

=> S₂ = 100².385

=> S₂ = 3850000

a, \(2A=8+2^3+2^4+...+2^{21}\)

\(\Rightarrow2A-A=2^{21}+8-\left(4+2^2\right)+\left(2^3-2^3\right)+...+\left(2^{20}-2^{20}\right)=2^{21}\)

b, (x+1)+(x+2)+(x+3)+...+(x+100)=5750 

= 100x + (1+2+...+100) = 5750 

=> x = [ 5750 - (1+2+...+100) ] :100 

=> x = (5750 - 5050) : 100 

=> x = 7

a: A=3^2(1^2+2^2+...+10^2)

=9*385

=3465

b: B=2^3(1^3+2^3+...+10^3)

=8*3025

=24200

giúp tui nha

thanks

giúp nha

A=4+2²+2³+2⁴+...+2²⁰

=2²+2²+2³+2⁴+...+2²⁰

=>2A=2³+2³+2⁴+...+2²¹

=>2A-A=2³+2³+2⁴+...+2²¹-2²-2²-2³-2⁴-...-2²⁰

=>A(2-1)=2³+2²¹-2²-2²

=>A=8+2²¹-4-4

=>A=2²¹

câu B làm thiếu bước rồi. Bạn làm tắt quá

b) ( x+1 ) + ( x+2 ) + ...+ ( x+100 ) = 5750

( x+x+x+...+x ) + ( 1+2+3+...+100 ) = 5750

( x.100 ) + (1.100) . 100 : 2 = 5750

( x.100 ) + 5050 = 5750

( x.100 ) = 5750-5050

 x.100 = 700

x = 700 : 100

x=7

Vậy x = 7.

a,2A=8+2³+2⁴+2⁵+............+2²¹

2A-A=(8+2³+2⁴+2⁵+...........+2²¹)-(4+2²+2³+2⁴+.............+2²⁰)

A=(8+2³+2²¹)-(4+2²)

A=8+8+2²¹-4-4

A=4+4+2²¹

A=2³+2²¹

b,(x+1)+(x+2)+.............+(x+100)=5750

=>x+1+x+2+...........+x+100=5750

=>100x+(1+2+............+100)=5750

=>100x+\(\frac{100.\left(100+1\right)}{2}\)=5750

=>100x+5050=5750

=>100x=5750-5050

=>100x=700

=>x=700:100

=>x=7

S=3⁰+3²+3⁴+3⁶+...+3²⁰⁰

6S=3²+3⁴+3⁶+...+3²⁰²

6S-S=(3²+3⁴+3⁶+...+3²⁰²)-(1+3²+3⁴+...+3²⁰⁰)

5S=1+(3²-3²)+(3⁴-3⁴)+...+(3²⁰⁰-3²⁰⁰)+3²⁰²

S=(3²⁰⁰+1):5\(\frac{ }{ }\)