Complete the proof in Example 11.4.4.

Complete the proof in Example 11.4.4.

Given information:

ak=2a⌊k/2⌋ for all integers k≥2a1=1

Proof:

To prove: an=2⌊log2n⌋ for all integers n≥1

PROOF BY STRONG INDUCTION:

Let P(n) be "an=2⌊log2n⌋"

Basis step: n = 1

a1=1=20=2⌊log21⌋

Thus P (1) is true.

INDUCTIVE STEP:

Let P(1),P(2),...,P(k) be true, thus ai=2⌊log2i⌋ for i=1,2,...,k

We need to prove that P ( k + 1) is true.

FIRST CASE: k odd

Since k is odd, k + 1 is even and thus ( k + 1) / 2 is an integer.

ak+1=2a⌊(k+1)/2⌋                                                     ak=2a⌊k/2⌋

=2a(k+1)/2                                      (k+1)/2 is an integer

=2⋅2⌊log2(k+1)/2⌋                                   P((k+1)/2) is true

=2⌊log2(k+1)/2⌋+1

=2⌊log2(k+1)−log22⌋+1                             logbxy=logbx−logby

=2⌊log2(k+1)−1⌋+1

=2⌊log2(k+1)−1+1⌋

=2⌊log2(k+1)⌋

SECOND CASE: k even

Since k is even, k + 1 is odd and thus k / 2 is an integer