Mathematical induction proves that 1 h n 1 1 hn is true for any n in the range of 0 h 1.

Mathematical induction proves that (1-h) n 1 (1+hn) holds for any natural number n in the range of 0 h 1.

Proof: n=1, 1-h) n=1-h

1/(1+hn)=1/(1+h)

by 0 h 1

1-h 1 (1+h), 1-h 1, is clearly true.

Assuming n=k, (1-h) k 1 (1+hk) holds.

then n=k+1.

1-h)^(k+1)=(1-h)^k(1-h)≤(1-h)/(1+hk)

If (1-h) (1+hk) 1 [1+h(k+1)] 1) then (1-h)[1+h(k+1)] 1+hk1+h+hk -h-h -kh 1+hkh +kh 0

by 0 h 1

Since h 0, k > 0, the above equation holds, i.e., hypothesis (1) holds.

According to the inductive method, (1-h) n 1 (1+hn) holds for any n in the range of 0 h 1.

This is obviously true when n=1, and it is still true when n=k, i.e., (1-h) k<=1-kh

When n=k+1, (1-h)(1-h) k<=(1-h)(1-kh)<(1-(k+1)h) is proved.

n=2 ((n+1)/2)^n= [2+1)/2]^2= n!=2*1=2 So((n+1) 2) n> n!Establish.

n>2 Suppose n=k holds the original formula, i.e., ((k+1) 2) k> k!i.e. (k+1) k 2 k>k!(1)

Then n=k+1, ((k+1+1) 2) (k+1)=(k+2) (k+1) (2*2 k).2)

Cause(k+2) (k+1)>2(k+1) (k+1).3)

3) Substituting (2) (k+1+1) 2) (k+1)=(k+2) (k+1) (2*2 k)>2(k+1) (k+1) (2*2 k)=(k+1) (k+1) 2 k=(k+1)*(k+1) k 2 k(4)

Substituting (1) into (4) gives ((k+1+1) 2) (k+1)>(k+1)*k!=(k+1)!i.e. n=k+1((n+1) 2) n > n!Establish.

This is true when n=1.

This is true when n=n.

then when n=n+1.

1＋2＋3＋…＋n+n+1

1/2）n*（n＋1）+n+1

1/2）(n+2)*（n＋1）

Established, so established.

Problem solving ideas: Directly use the mathematical induction method to prove the steps of the problem, and prove the inequality can be proved: (1) When n=2, [1 2+1+

24] The proposition holds

2) Suppose that when n=k, [1 k+1+

k+2+k+3+…+

2k 24] was founded.

When n=k+1, [1 k+2+

k+3+…+

2k+2k+1+

2k+2]=[1/k+1]+[1/k+2+k+3+…+

2k+2k+1+

2k+2]−

k+1＞2k+1+

2k+2−k+1，∵[1/2k+1+

2k+2−k+1＝

2(2k+1)(k+1)＞0，∴

k+1)+1+

k+1)+2+

k+1)+3+…+

2(k+1)＞

24], when n=k+1 the proposition holds

So for any n 2, n n* holds , 10,

Proof 1: When n=1, the left side = 1*1!=1, right =(1+1)!-1=2-1=1

i.e. left = right.

2. The assumption that n=k(k 1) is the conclusion is true.

i.e. 1 1!+2•2!+.k•k!=(k+1)!-1 then when n=k+1, 1 1!+2•2!+.k•k!+（k+1）（k+1）!

k+1)!-1+（k+1）（k+1）!

k+1)!+k+1）（k+1）!-1

1+（k+1）]（k+1）!-1

k+2）（k+1）!-1

k+2）!-1

k+1+1）!-1

That is, the conclusion of n=k+1 is true.

Therefore, in summary, the original proposition is valid.

When n=1, 1*(1+1)(2*1+1) 6=1, holds.

When n=2, 2*(2+1)(2*2+1) 6=5, holds.

Assuming n=k, 1 + 2 + 3 + ...n = n(n+1)(2n+1) 6 is true, but when n = k+1, 1 +2 +3 +....k²+(k+1)²=(1²+2²+3²+…k²)+k+1)²=k（k+1）（2k+1）/6+(k+1)²=(k+1)(k(2k+1)/6+(k+1))=(k+1)((2k²+k+6k+6)/6)=(k+1)(2k²+k+6k+6)/6

k+1)(2k²+7k+6)/6

k+1)((k+2)(2k+3))/6=(k+1)(k+2)(2k+3)/6

k+1)((k+1)+1)(2(k+1)+1) 6.

Proof that 1' when n=1, left = (2*1-1) =1, right = 1 3*1*(4*1 -1), left = right, the proposition is true.

2' Suppose the proposition holds when n=k.

i.e. 1 + 3 + 5 +2k-1) = 1 3k (4k-1) holds.

Then when n=k+1, left=1 +3 +5+2k-1)²+2(k+1)-1]²=1/3k(4k²-1)+(2k+1)²

1/3k+4k²+4k+1

4/3k3+4k²+11/3k+1

Right = 1 3 (k + 1) [4 (k + 1) -1] = (1 3k + 1 3) ( 4k +8k + 3) = 4 3k 3 + 8 3k + 8 3k + 1

4/3k3+4k²+11/3k+1

Left = Right, the proposition is true.

The proposition from 'is that any n belongs to n*.