Use Mathematical Induction to prove that for [; n>=1 ;], that $ b_n=(1/2)((3^n)+1) $
Solution:
Basic case: For [; n = 1 ;] $$ b_1 = 2 = 1/2((3^1)+1) $$
Assume that for some k $$ b_k = 1/2(3^k + 1) $$
Then $$ b_{k+1} = 3b_k - 1 $$ $$ = 3(1/2)(3^k + 1) - 1 $$ $$ = 3/2(3^k) + 3/2 - 1 $$ $$ = 3/2(3^k) + 1/2 $$ $$ = 1/2(3^{k+1} + 1 $$
I just want to figure out how exactly we got $ b_{k+1} = 3b_k - 1 $ in the first place? What has led us to that assumption or where did it come from (I understand everything else)?
In the problem statement, there should be $b_1=2$ and $b_{n+1}=3b_n-1$. Without those, you don't have enough information to solve the problem.