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

http://olm.vn/hoi-dap/question/223617.html. có bài giống

A=1+3/2^3+4/2^4+5/2^5+...100/2^100
1/2*A = 1/2 + 3/2^4 + 4/2^5 +....+ 99/2^100 + 100/2^101

A- A/2 = 1/2A =1/2 + 3/2^3 + 1/2^4 +...+1/2^100 - 100/2^101=

= [1/2+1/2^2 +1/2^3 +...+1/2^100] -100/2^101 (Do 3/2^3 = 1/2^2 +1/2^3)

=[1-(1/2)^101]/(1-1/2) -100/2^101 =

=(2^101 -1)/2^100 - 100/2^101

=> A= (2^101 -1)/2^99 - 100/2^100

A=1+3/2^3+4/2^4+5/2^5+...100/2^100 
1/2*A = 1/2 + 3/2^4 + 4/2^5 +....+ 99/2^100 + 100/2^101 

A- A/2 = 1/2A =1/2 + 3/2^3 + 1/2^4 +...+1/2^100 - 100/2^101= 

= [1/2+1/2^2 +1/2^3 +...+1/2^100] -100/2^101 (Do 3/2^3 = 1/2^2 +1/2^3) 

=[1-(1/2)^101]/(1-1/2) -100/2^101 = 

=(2^101 -1)/2^100 - 100/2^101 

=> A= (2^101 -1)/2^99 - 100/2^100 

A=1+3/2^3+4/2^4+5/2^5+...100/2^100 
1/2*A = 1/2 + 3/2^4 + 4/2^5 +....+ 99/2^100 + 100/2^101 

A- A/2 = 1/2A =1/2 + 3/2^3 + 1/2^4 +...+1/2^100 - 100/2^101= 

= [1/2+1/2^2 +1/2^3 +...+1/2^100] -100/2^101 (Do 3/2^3 = 1/2^2 +1/2^3) 

=[1-(1/2)^101]/(1-1/2) -100/2^101 = 

=(2^101 -1)/2^100 - 100/2^101 

=> A= (2^101 -1)/2^99 - 100/2^100 
 

giải thích cho minh 3 dong cuoi 

3850000