Math 213, Section B1, Quiz 2 (SOLUTIONS); Friday, February 2, 2007
1. Using induction, prove that for every integer n≥2
3n≥2n2+ 1.
Solution. 1) Base of Induction. First, check if the statement 3n≥
2n2+ 1 holds for n= 2.
We have 32= 9 and 2 ·22+ 1 = 2 ·4 + 1 = 9. Since 9 ≥9, the
required statement does hold for n= 2.
2) Inductive Step.
Let k≥2 and suppose that 3k≥2k2+ 1 is known to hold. We need
to derive that 3k+1 ≥2(k+ 1)2+ 1, that is, 3k+1 ≥2(k2+ 2k+ 1) + 1,
that is, 3k+1 ≥2k2+ 4k+ 3.
The inductive hypothesis
3k≥2k2+ 1 by multiplying by 3 implies
3k+1 ≥6k2+ 3.
To show that 3k+1 ≥2k2+ 4k+ 3 it suffices to establish that for k≥2
we have 6k2+ 3 ≥2k2+ 4k+ 3.
We have:
6k2+ 3 ≥2k2+ 4k+ 3 is equivalent to:
4k2≥4kby dividing by 4k > 0, is equivalent to:
k≥1,
which holds since by assumption k≥2. Thus for k≥2 we have
6k2+ 3 ≥2k2+ 4k+ 3, as required.
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