Verify that n= 1, 2, 3, 4, ... or with the help of images with functions y=2x and y=x2, find out that the smallest positive integer m is equal to 6, and then prove it according to the steps of mathematical induction.

Solution: 2n- 1 < (n+ 1) 2 when n= 1.

When n=2, 22- 1 = 2 < (2+ 1) 2.

When n=3, 23- 1 = 4 < (3+ 1) 2.

When n=4, 24- 1 < (4+ 1) 2.

When n=5, 25- 1 < (5+ 1) 2.

When n=6, 26- 1 < (6+ 1) 2.

When n=7, 27- 1=(7+ 1)2.

When n=8, 28- 1 > 8+ 1) 2.

…

When n≥8, 2n- 1 > (n+ 1) 2 is constant, the conjecture holds.

Mathematical induction proves that:

(1) When n=8, 28- 1= 128, (8+ 1) 2 = 8 1, 128 > 8 1

(2) suppose that the inequality holds when n=k(k≥8), that is, there is 2k- 1 > (k+ 1) 2.

Then when n=k+ 1, 2(k+ 1)- 1=2k=2? 2k- 1>2？ (k+ 1)2 = k2+[(k+2)2-2]>(k+2)2(∵k2-2 > 0)

=[(k+ 1)+ 1]2, that is, when n=k+ 1, the inequality also holds.

According to (1)(2), 2n- 1 > (n+ 1) 2 holds when n≥8.