Proof by Induction: Prove a(n)=2^(n-1)[n+2]

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SUMMARY

The discussion focuses on proving the formula a(n) = 2^(n-1)[n+2] for the sequence defined by a(0) = 1, a(1) = 3, and a(n) = 4[a(n-1) - a(n-2)] for n ≥ 2 using strong induction. The initial step involves verifying the base case for n = 0. The proof then utilizes the assumption that the formula holds for all k less than n, specifically for k = n-1 and k = n-2, to derive a(n) from the recurrence relation.

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Homework Statement


Define the numbers a(0),a(1),a(2),...by a(0)=1, a(1)=3, a(n)=4[a(n-1) - a(n-2)]. for n>=2
Show by induction that for all n>=1, a(n)=2^(n-1) [n+2]

Homework Equations


The Attempt at a Solution



proof by induction is not my stong point, and i really don't know where to start on this question. Any help would be much appreciated!
 
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One nice thing about induction is you always know "where to start"- prove the statement is true for n= 0!

Then you will really want to use "strong induction": If, whenever a statement is true for all k less than n, it true for k= n, then the statement is true for all positive integers.

If the statement is true for all k< n, it is, in particular true for n-1 and n-2: you can assume that
a(n-1)= 2^{n-1-1}(n-1+ 2)= 2^{n-2}(n+1)
and
a(n-2)= 2^{n-2-1}(n-2+2)= 2^{n-3}n

Put those into a(n)= 4[a(n-1)- a(n-2)] and see what you get.
 


thanks for the help!
 

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