Question

Suppose F(n+1)=2F(n)+1/2 for n=1,2,3....... and f(1)=2 then f(101)n is how much?

Answer 1

F(n+1) = 2 F(n) + 1/2    for n = 1,2,3...

F(1) = 2
F(2) = 2 * 2 + 1/2 = (2^3+1)/2 = 2^2 + 1/2 = 9/2 = (10 -1)/2
F(3) = 2 [2^2+1/2] +1/2 = (2^3+1) +1/2 = (2^4+2+1)/2 =19/2
       =(20-1)/2 = (2^2*5-1)/2
F(4) = 2 [2^4+2+1] +1/2=[2^6+2^3+2^2+1]/2 = 39/2=(40-1)/2 =(2^3*5-1)/2
F(5) = (80-1)/2 = (2^4 * 5 -1)/2
F(6) = 80-1/2 = (160-1)/2 = (2^5* 5 - 1)/2

F(n+1) = [5 * 2^(n-1) - 1] / 2     
  
So  F(101) = [ 5 * 2^100 - 1] / 2  = [10 * 2^99 - 1 ] / 2
                  =  10 * 2^98 - 1/2
                   = 3169126500570573503741758013439.5