For any positive integer, guess the magnitude relationship of 2 n 2 and n 2 2 and prove 10

n=1 at 2 3<3 2

n=2 when 2 4=4 2

n = 3 at 2 5> 5 2

n = 4 at 2 6 > 6 2

Therefore, suppose that when n>2, 2 (n 2) > (n 2) 2 is proved by mathematical induction.

The assumption is true when n=n.

i.e. 2 (n 2) > (n 2) 2

then when n=n+1.

2^(n+1＋2)-（n+1＋2）^2

2*2^(n＋2)-（n＋2）^2-2(n+2)-12^(n＋2)-2(n+2)-1

2^(n＋2)-（n＋2）^2

Hence the conjecture.

When n=1, 2 (1+2)<(1+2) 2When n=2, 2 (2+2)=(2+2) 2 When n=3, 2 (2+3)=32>(3+2) 2=25 Conjecture 2 (n+2)>(n+2) 2, proved by mathematical induction.

When n=3, 2 (2+3)=32>(3+2) 2=25 Suppose that when n=k, 2 (k+2)>(k+2) 2 when n=k+1.

2 (k+1+2)=2 (k+2)*2>2*(k+2) 2k+1+2) 2=(k+2) 2+1+2(k+2) (k+2) (k+2) >2-1-2(k+2)=k 2+4k+4-1-4k-4=k 2-1>0

So 2*(k+2) 2>(k+1+2) 2 i.e. when n=k+1, 2 (k+1+2)>(k+1+2) 2, so when n>=3, 2 (n+2)>(n+2) 2 is proven.

This gives us the respect for any positive integer.

When n=1, 2 (n+2)<(n+2) 2 When n=2, 2 (n+2)=(n+2) 2 When n>=3, 2 (n+2)>(n+2) 2

Mathematical Induction:

n=1 at 2 3<3 2

n=2 when 2 4=4 2

n = 3 at 2 5> 5 2

n = 4 at 2 6 > 6 2

Therefore, suppose that when n>2, 2 (n 2) > (n 2) 2 is proved by mathematical induction.

The assumption is true when n=n.

i.e. 2 (n 2) > (n 2) 2

then when n=n+1.

2^(n+1＋2)-（n+1＋2）^2

2*2^(n＋2)-（n＋2）^2-2(n+2)-12^(n＋2)-2(n+2)-1

2^(n＋2)-（n＋2）^2

Hence the conjecture.

Or: make use of images.

<> can be seen from the diagram of the upper trapped auspicious answer, the integer between -2 and 2 has ;

Therefore, the answer to the banquet is Wang Hui.

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

When n=2, 2 n-1 = 2, (n+1)2 = 9,2 n-1<(n+1) 2

When n=3, 2 n-1 = 4, (n+1)2 = 16,2 n-1<(n+1)2

n=4,2 n-1 = 8,(n+1)2 = 25,2 n-1<(n+1)2

When n=5, 2 n-1 = 16, (n+1)2 = 36,2 n-1<(n+1) 2

n=6, 2 n-1 = 32, (n+1)2 = 49,2 n-1<(n+1)2

At n=7, 2 n-1 = 64, (n+1) 2 = 64, 2 n-1=(n+1) 2

n=8,2 n-1 = 128,(n+1)2 = 81,2 n-1>(n+1)2

Suppose n=k, 2 k-1>(k+1) 2

When n=k+1, 2 k+1-1 = 2*(2 k-1).

2*（k+1)^2 =（k+1)^2 + k+1)*(k+1)

k+1) 2 + k*(k+1) + k+1 (because k>2).

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

k+1+1)^2

So, by mathematical induction, it is proved that when n>=8, there are 2 n-1 >(n+1) 2

Using mathematical induction, we get (n+1) 2>2 (n-1).

You can start by making a table and taking the specific value of n, comparing the size of 2 n and n 2, and find that when n=1 has 2 n>n 2

n=2 or 4 equal.

n=3 has 2 n4 and has 2 n>n 2

The proof can be forged by mathematical induction.

Solution: When n 3, 2 3, 2 n n

When n 2 and n 4, 2 n n

When n is another positive integer, 2 n n

There are right angles.

And Clibi Rock 3 = Fierce B

So for the triangle Hui Zhao.

Yes edf= dgb

So de bc

So aed= c

2 (n-1) increases to (n+1) 2

Fast, it is easy to see from the law of Robida in the rock. , so when n keeps increasing, 2 (n-1) must be greater than (n+1) 2When the calculator is used to test and get n to take 1 7.

2 (n-1)=8.

2 (n-1)>(n+1) But since you say it's a positive integer, I'll use mathematical induction to prove the following proposition: at n>=8.

2ˆ（n-1)>（n+1）²1

When n=8, the proposition holds 2

The proposition is true when it is assumed that n=k.

i.e. 2 (k-1) > (k+1).

Then when n=k+1

time 2*k=2*

2*(k-1))> Caution2(k+2) 2>(k+1) 2The last inequality is good after you, and the code is respectful.

Answer: 1+2+2 2+2 3+2 4=31;

1+2+2^2+..2 14 = 2 15-1 is maximally divisible by 31.

Proof 1+2+2 2+.2^(5n-1)=2^(5n)-1

32^n-1

31+1)^n-1

31k+1-1

31k 1+2+2^2+..2 (5n-1) is maximally divisible by 31.