Giving the recurrence f( n )=6f(n-1)-9f(n-2) for n>=3. The root is unique ...
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Giving the recurrence f( n )=6f(n-1)-9f(n-2) for n>=3. The root is unique which is 3. The constants c1=7/9 and c2=-1/9. Therefore, f( n ) is:
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