Select the solution to each of these recurrence relations: 2(n + 1)! 5. 3· 3n + 1 (-1)^n (2^n)n! an = 2nan-1, a0 = 3 %3D an = O O O -an-1, a0 = 5 an = an-1 + 3, а0 %D 1 an = (n %3D + O O O 1)an-1, a0 = 2

Q: - f(n) = log n²; g(n) = log n + 5 - f(n) = n; g(n) = log n? - f(n) = log log n; g(n) = log n f(n) =…

A: analysis for the above example is in step 2.

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A:

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A:

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A: Hi please find the solution below and I hope it would be helpful for you. According to guidelines I…

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A:

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A: The big-O of the given functions

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A: The above question is solved in step 2 :-

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A: The above question is solved in step 2 :-

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A: Below I have provided the solution of the given question

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A: Note: As, per our guidelines we can answer only three subdivisions, So please repost the remaining…

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A: The handwritten answer is below:

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A: 1. Replace n+1 by n in the recurrence relation. 2. Substitute n=1,2,3...n to get S1, S2, S3,....,Sn.…