(1), n=5, obviously established.

(2)

If n=k is true, k >；; =5

Namely k 2

Then n=k+ 1.

(k+ 1)^2=k^2+2k+ 1<； 2^k+2k+ 1

2^k+2k+ 1-2^(k+ 1)=2^k+2k+ 1-2*2^k

=2k+ 1-2^k

k^2-(2k+ 1)=k^2-2k- 1=(k- 1)^2-2

Because k & gt=5

So (k- 1) 2-2 >: 0

So k 2 > 2k+ 1

So 2k > k 2 > 2k+ 1

So 2k+ 1-2 k < 0.

So (k+ 1) 2

Comprehensive (1), (2)

Proof of proposition