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Consider the following recurrence relation.

B(n) = 2 if n=1

B(n)= 3*B(n − 1) + 2  if n > 1


Use induction to prove that B(n) = 3n − 1.

(Induction on n.) Let f(n) = 3n − 1.



Base Case: If n = 1,

 the recurrence relation says that B(1) = 2,

 and the formula says that f(1) = 3 − 1 =  ,

 so they match.


Inductive Hypothesis: Suppose as inductive hypothesis that B(k − 1) =  

 for some k > 1.



Inductive Step: Using the recurrence relation,

B(k) = 3 · B(k − 1) + 2, by the second part of the recurrence relation

 = 3    + 2, by inductive hypothesis

 = (3k − 3) + 2

 =   



so, by induction, B(n) = f(n)

 for all n ≥ 1.

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