Ta có :
\(S=1+3+3^2+....+3^{2014}\)
\(\Rightarrow\left(3-1\right)A=\left(3-1\right)1+\left(3-1\right)3+\left(3-1\right)3^2+....+\left(3-1\right)3^{2014}\)
\(\Rightarrow2A=3-1+3-3^2+....+3^{2015}-3^{2014}\)
\(\Rightarrow2A=3^{2015}-1\)
\(\Rightarrow2B-2A=3^{2015}-\left(3^{2015}-1\right)\)
\(\Rightarrow2B-2A=1\)
\(\Rightarrow2\left(B-A\right)=1\)
\(\Rightarrow B-A=\frac{1}{2}\)
S = 1 + 3 + 32 + ... + 32014
= > ( 3 - 1 ) A = ( 3 - 1 ) 1 + ( 3 - 1 ) 3 + ( 3 - 1 ) 32 + ... + ( 3 - 1 ) 32014
= > 2A = 3 - 1 + 3 - 32 + ... + 32015 - 32014
= > 2A = 32015 - 1
= > 2B - 2A = 32015 - ( 32015 - 1 )
= > 2B - 2A = 1
= > 2 ( B - A ) = 1
= > B - A = \(\frac{1}{2}\)
Vậy B - A = \(\frac{1}{2}\)
A = 1+3+3^2 + 3^3 + ......+3^2014
Ta thấy các số hạng liền sau gấp số hạng liền trước 3 lần
A.3=3( 1+3+3^2 + 3^3 + ......+3^2014)
A.3= 3+3^2+3^3+3^4+........+3^2015
A.3-A=(3+3^2+3^3+3^4+........+3^2015) - (1+3+3^2 + 3^3 + ......+3^2014)
A.2= 3 ^2015 - 1
B.2=3^2015
=> B.2-A.2= 3^2015 - (3^2015 - 1 )
=> B-A = 1