There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n is a positive integer. What is the value of A(66)-A(65)?  Choices.
A) 2^22+1  B) 2^65  C) 2^23+1  D) 2^24  E) 2^23+1
A1= 3   A2= 2*3-1= 
<gadget id="calculator">2*3-1</gadget>
<output>5</output>
5   A3= 2*5-1= 
<gadget id="calculator">2*5-1</gadget>
<output>9</output>
9   A4= 2*9-1= 
<gadget id="calculator">2*9-1</gadget>
<output>17</output>
17   We can notice that there is a squence   A2-A1= 2^1   A3-A2= 2^2   A4-A3= 2^3   Hence A66-A65= 2^65   2^65 is the answer
<result>B</result>
There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n is a positive integer. What is the value of A(66)-A(65)?