Ta có : A = 1 + 3 + 32 + ....... + 32015
=> 3A = 3 + 32 + ....... + 32016
=> 3A - A = 32016 - 1
=> 2A = 32016 - 1
=> A = 32016 - 1/2
Mà B = 32016/2
=> B - A = 32016/2 - 32016 - 1/2
=> B - A = 1/2
Ta có: 3a-a=32016-1 <=> a=\(\frac{3^{2016}-1}{2}\)
=> b-a=\(\frac{3^{2016}}{2}-\frac{3^{2016}-1}{2}=\frac{1}{2}\)
a= 1+ 3 + 3 + 3^2 +.... + 3^2015
=> 3a = 3+ 3^2 + 3^3 +..... + 3^2016
=> 3a - a = (3+ 3^2 + 3^3 +.... + 3^2016) - ( 1 + 3+ 3^2 +......+ 3^ 2015)
=> 2a = 3^2016 - 1
=> a= \(\frac{3^{2016}-1}{2}\)< b= \(\frac{3^{2016}}{2}\)
vậy a< b