Question

a, A=4+4^2+4^3+......+4^n

b, B=1+3+3^2+....+3^100

Answer 1

a, A = 4 + 4^2 + 4^3 + ... + 4^n 

=> 4A = 4.(4 + 4^2 + 4^3 + ... + 4^n) 

=> 4A = 4^2 + 4^3 + 4^4 + ... + 4^n+1 

=> 3A = 4A - A = (4^2 + 4^3 + 4^4 + ... + 4^n+1) - ( 4 + 4^2 + 4^3 + ... + 4^n) 

=> 3A  = 4^n+1 - 4 

=> A    = \(\frac{4^{n+1}-4}{3}\)

Vậy A = ..................

b, B = 1 + 3 + 3^2 + ... + 3^100 

=> 3B = 3.(1 + 3 + 3^2 + ... + 3^100) 

=> 3B = 3 + 3^2 + 3^3 + ... + 3^101 

=> 2B = 3B - B =(3 + 3^2 + 3^3 + ... + 3^101) -(1 + 3 + 3^2 + ... + 3^100)

=> 2B = 3^101 - 1 

=> B = \(\frac{3^{101}-1}{2}\)

Vậy B = ...................... 

Answer 2

  A = 4 + 4^2 + 4^3 + ... + 4^n

4A = 4^2 + 4^3 + 4^4 + ... + 4^n+1

4A - A = ( 4^2 + 4^3 + 4^4 + ... + 4^n+1 ) - ( 4 + 4^2 + 4^3 +...+4^n)

3A = 4^n+1 - 4

  A = 4^n+1 - 4/3

Answer 3

B cx là cách đó