f 1 2, f n 1 f n f n 1, where n belongs to the set of positive integers. Verify 1 f

From the known f(n+1)=[f(n)] 2-f(n)+1, we know that f(n+1)-1=f(n)[f(n)-1].

and f(1)=2, we know that f(n) is an increasing function, and.

1/[f(n+1)-1]=1/[f(n)(f(n)-1)]=1/[f(n-1)-1]-1/f(n).

1/f(n)=1/[f(n)-1]-1/[f(n+1)-1].

Replace n = 1, 2, 3 ,..n is substituted into the above equation and added to give <1, n>[1 f(k)]=1 [f(1)-1]-1 [f(n+1)-1]<1

For the positive integer n 2, there is.

1/f(1)+1/f(2)+1/f(3)+.1/f(n)<1.

If you are satisfied!! Thank you.

It should be to prove.

1 - 1 2 2 < 1 - 1 2 2 n, because if you follow the original question, f(1) = 2, f(2) = 3, 1 2+1 3 = 5 6, and 1-1 2 = 7 8 > 5 6, which is not in line with the question.

It is known by f(n+1) -1 = (f(n) -1) *f(n).

1/[f(n+1)-1] = 1/[f(n)-1] -1/f(n)

So 1 f(i) = 1 [f(i)-1] -1 [f(i+1)-1].

For i to be summed from 1 to n, the right side of the above equation can cancel each other out, as long as it is proved.

1-1/2^2^ <1 - 1/[f(n+1)-1] <1 - 1/2^2^n

This is equivalent to: 2 2 The above equation can be proved by mathematical induction. First, when n = 2, f(3) = 7, ie.

2 2< 7 - 1 < 2 4 = 16, the conclusion is true.

If n is assumed that the inductive assumption is true, then.

1)2^2^ 2^2^ +1) *2^2^ >2^2^n

2) f(n+1) -1 < 2 2 n, then f(n+1) < = 2 2 n (because f(n) can only be taken as an integer).

f(n+2) -1 = f(n+1)(f(n+1) -1) <2^2^n * 2^2^n = 2^2^

This completes the inductive proof.

When f(1)=1, f(2)=1, n>2, f(n)=f(n-1)+f(n2). According to this, it can be deduced that when n>1, there is a recursive relation of the vector: (f(n+1),f(n))=f(n),f(n 蚂ye Zheng 1)*a.

where a is a matrix of 2*2 ( ?) thus, (f(n+1),f(n)))=f(2),f(1))*

According to the recursive relation f(n) =f(n-1) +f(n-2) given in the question, we can write it as a matrix: [f(n+1)] 1 1] [f(n)][f(n)] 1 0] *f(n-1)] where [1 1] and [1 0] are 2*2 guess matrices, respectively. We can write it as a:

a = 1 1] [1 0] So, according to the recursive relation of the question, we can get: [f(n+1), f(n)] f(n), f(n-1)] a where, (f(2), f(1)) 1, 1). So, (f(n+1), f(n)) 1, 1) *a.

f(n+1)-f(n)=1 2, so.

f(2)-f(1)=1/2

f(3)-f(2)=1/2

f(4)-f(3)=1/2

f(101)-f(100)=1/2

The above is added to obtain: f(101)-f(1)=100*1 2=50So, f(101)=52

1) by f(m+n) = f(m) + f(n) + 4 (m + n) - 2

then f(n)f(n-1+1).

f(n-1)+f(1)+4n-2

f(n-1)+4n-1

f(n-2)+4(n-1)-1+4n-1

f(1)+4*1+4*2+……4(n-1)+4n-(n-1)

1+4n(n-1)/2-n+12n^2-3n+2

2n^2-3n+2

then f(x)=2x 2-3x+2,(x n+).

2) Let g(m)=m 2-tm-1, then only g(m)max<=f(x)min is needed to satisfy m 2-tm-1 f(x).

Then the axis of symmetry of f(x) is x=3 4, then f(x) is an increasing function in x n+, then f(x)min=f(1)=1

The axis of symmetry of g(m) m=t 2, when t<=-2, t 2<=-1, g(m) is the increasing function, then g(m)max=g(1)=-t-1, there is no solution.

When t>=2, t 2>=1, g(m) is the subtraction function, then g(m)max=g(-1)=t=t>-2, g(m)max=g(1)=-t-1, then -1f(m+2).

2*（m+2）^2-3*(m+2)+2

The above inequality is reduced to.

1/3*m^3-2*m^2-16/3*m-5<0

Let g(m)=1 3*m 3-2*m 2-16 3*m-5, then the derivative g'(m) = m 2-4 * m - 16 3, when m > = 6, g'(m)>0, g(m) is the increasing function, g(m)min=g(6)=36-24-16 3=20 3>0, then g(m)>0 is constant, which does not meet the topic;

When 2<=m<=5, g'(m)<0, g(m) is the subtraction function, g(m)max=g(5)=25-20-16 3=-1 3<0, then g(m)<0 is constant, which is in line with the topic.

Therefore, the maximum value of m is 5.

It should be to prove.

1 - 1 2 2 < 1 - 1 2 2 n, because if you follow the original question, f(1) = 2, f(2) = 3, 1 2+1 3 = 5 6, and 1-1 2 = 7 8 > 5 6, which is not in line with the question.

It is known by f(n+1) -1 = (f(n) -1) *f(n).

1/[f(n+1)-1] = 1/[f(n)-1] -1/f(n)

So 1 f(i) = 1 [f(i)-1] -1 [f(i+1)-1].

For i to be summed from 1 to n, the right side of the above equation can cancel each other out, as long as it is proved.

1-1/2^2^ <1 - 1/[f(n+1)-1] <1 - 1/2^2^n

This is equivalent to: 2 2 The above equation can be proved by mathematical induction. First, when n = 2, f(3) = 7, ie.

2 2< 7 - 1 < 2 4 = 16, the conclusion is true.

If n is assumed that the inductive assumption is true, then.

1)2^2^ 2^2^ +1) *2^2^ >2^2^n

2) f(n+1) -1 < 2 2 n, then f(n+1) < = 2 2 n (because f(n) can only be taken as an integer).

f(n+2) -1 = f(n+1)(f(n+1) -1) <2^2^n * 2^2^n = 2^2^

This completes the inductive proof.

From the known f(n+1)=[f(n)] 2-f(n)+1, we know that f(n+1)-1=f(n)[f(n)-1].

and f(1)=2, we know that f(n) is an increasing function, and.

1/[f(n+1)-1]=1/[f(n)(f(n)-1)]=1/[f(n-1)-1]-1/f(n).

1/f(n)=1/[f(n)-1]-1/[f(n+1)-1].

Replace n = 1, 2, 3 ,..n is substituted into the above equation and added to give <1, n>[1 f(k)]=1 [f(1)-1]-1 [f(n+1)-1]<1

For the positive integer n 2, there is.

1/f(1)+1/f(2)+1/f(3)+.1/f(n)<1.

1：f(n)-f(n-1)=-1 f(n-1)-f(n-2)=-1 f(n-2)-f(n-3)=-1 ；；f(1)-f(0)=-1

F(n)=3-n f(4)=3-4=-12: Suppose the median range is n-1 in the interval

So the range of y=[x] is an integer.

4: As long as the axis of symmetry 1-a 4 is solved as a -3

(1): f(0)= So.

f(1)-f(0)=-1 .f(1)=2

f(2)-f(1)=-1 .f(2)=1

f(3)-f(2)=-1 .f(3)=0

f(4)-f(3)=-1 .f(4)=-1

2): It can be known by observation.

f(1)-f(0)=-1 .f(1)=2

f(2)-f(1)=-1 .f(2)=1

f(3)-f(2)=-1 .f(3)=0

f(4)-f(3)=-1 .f(4)=-1

f(n)-f(n-1)=-1 is added to both sides of the equation.

f(1)-f(0)+f(2)-f(1)..f(n)-f(n-1)=-n

That is, f(n)=3-n; This number must be smaller than 2 and is an integer.

So the range of y is y<=2.

3): The key to this kind of problem is to remove the absolute value. Divide x>0 and x<0 to remove the absolute value.

When x>0, x -2|x|=a-1 equivalent x 2-2x+1=a

The left side is a perfectly flat method, (x-1) 2=a

When a<0, the equation has no solution. When a=0, the equation has a solution x=1 a>0 The equation has two solutions.

When x>0, x -2|x|=a-1 equivalent x 2-2x+1=a

The left side is a perfectly flat method, (x-1) 2=a

When a<0, the equation has no solution. When a=0, the equation has a solution x=1 a>0 The equation has two solutions.

4): The function f(x)=x +2(a-1)x+2 is a subtraction function over the interval (- 4).

Explain that any x is smaller than 4, then f(x)-f(4)>0 is .

x^2+2(a-1)x+2-4^2-2(a-1)*4-2>0

x^2+2(a-1)x-8(a+1)>0

Use the universal formula to solve the solution of x 2 + 2 (a-1) x - 8 (a + 1) = 0 and use the "number axis through the needle" method.

Next, you can do it yourself.

1.-1

(a=0 or a>1).

4 (1>a>0).

0 (a<0).

3 (a=1).

4.(-infinity, -3].

.It should be right...