The least n for which the recursive formula is applicable is n=1, for the closed form is n=0. When we go from k to k+1 we deal with a formula

a(k+1) = a(k) + 2*a(k-1).

This recursive formula holds from k>=1, so k+1>=2 and the first n for which such a proof is possible is n=2. So we ought to check the direct form for n=1 and n=0.

The answer is 2 (base cases).

Actually, they are not true (a0 = 1 not= 2^0+(-1)^0 = 2, a1 = 3 not= 2^1+(-1)^1 = 1).

Apr 30th, 2015

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