Math proof problems, math problems need to be proved because so

General term 1 n(n+1)(n+2).

1/n-1/n+1](1/n+1)

1/n(n+1)-1/(n+1)(n+2)

1/2(1/n-1/n+2)-1/n+1+1/n+2

1/2[1/n+1/n+2-2/n+1]

1/2[1/n-1/n+1+1/n+2-1/n+1]

So 1 1*2*3 +1 2*3*4+.1/n(n+1)(n+2)

1/2(1-1/2+1/3-1/2+1/2-1/3+1/4-1/3+1/3-1/4+1/5-1/4+..1/n-1/n+1+1/n+2-1/n+1)

1/2[1-1/2+1/(n+1)(n+2)]

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

Because 1 2(n+1)(n+2)>0 so -1 2(n+1)(n+2)<0

So 1 1*2*3 +1 2*3*4+.1/n(n+1)(n+2)<1/4=

o(∩_o...

It can be proved by mathematical induction as follows:

1) When n=1, left = 1 6 < 4, the inequality holds.

2) Assuming that n=k is true, that is.

1/6 < 1/1*2*3+1/2*3*4+..1 k(k+1)(k+2) <4, then, when n=k+1, there is.

1/1*2*3+1/2*3*4+..1/k(k+1)(k+2)》+1/(k+1)(k+2)(k+3)

And through simple proof, we can know that 1 (k+1)(k+2)(k+3) <1 24, so "1 1*2*3+1 2*3*4+.."1/k(k+1)(k+2)》+1/(k+1)(k+2)(k+3) <4

This means that the inequality also holds when n=k+1.

According to (1) and (2), it is known that the equation holds for any positive integer n.

So 1 1*2*3 +1 2*3*4+.1/n(n+1)(n+2)＜4

Proof questions, when there is a rock shooting, will be easier than solving the question.

Because it tells you the result, it just lets you explain the process.

First of all, you should be familiar with the relevant formulas of such questions and have certain experience in doing them.

You should be able to see the questions to know what knowledge points the questioner is going to test.

Secondly, for topics that have not been "seen", you can consider them together from both sides and push them together.

Again, familiarize yourself with the relevant problem-solving techniques. This also requires experience, and when you see the question, you will know what method to prove it is better to use Rolling Chang.

In addition, I personally suggest that if you can't think of it, you can try a few special values to see the process. Maybe it will be of great help.

1. Let the ab side pure cover wisdom be x, and Hu will answer ad=2x-2, so 5x-3(2x-2)=1

So x=5 is a parallelogram, so the perimeter is 262, and the solution: let de=2k

df=3kab=x

then bc=50-x

According to the area formula.

2k*x=(50-x)*3k

2x=3(50-x)

x=3050-x=20

ab=cd=30cm,bc

cd=20cm

x1+x2+x3=constant c,y=kx+b (k,b,x1,x2,x3 are all positive).

x1*y1+x2*y2+x3*y3

x1*(kx1+b)+x2*(kx2+b)+x3*(kx3+b)=k(x1^2+x2^2+x3^2)+b(x1+x2+x3)>=k(x1+x2+x3)^2/3+b(x1+x2+x3)=kc^2/3+bc

If and only if x1=x2=x3 is equal to x1, so when x1=x2=x3, the value of x1*y1+x2*y2+x3*y3 is the minimum value kc 2 3+bc

It's too late today, I'll tell you tomorrow morning.

First train of thought:

First of all, you can take the original formula, f(x) = (x+a 2) squared + (b-(a squared 4)).

The counter-argument method is to assume that m, i.e., the maximum absolute value of f(x) can be less than, and the application of the counter-argument method is to show that the maximum value will be smaller. After the analysis of the simplified formula, the positive value greater than 0 is in the front parentheses, and when the latter part is positive, it will not be the smallest, so when considering the latter is positive, it must be = 0 to reach the extremum. That is, only consider the previous ones, or discuss the extreme values, let x=1, -1, 0 respectively, calculate the maximum value, you will find that they are all greater than , this process can be seen in the above idea of using images.

There is also a case that the latter part is negative, in the same way, if you want the absolute value to be as small as possible, the latter should not be too large (because it is an absolute value) or equal to 0, the other steps are the same, and the final conclusion is that your original hypothesis is not true, and it is proved.

The second way of thinking:

You go to the original formula, |f（x）|+|, squared of =(x+a2).(b-(a's square 4))|Notice the symbol for absolute values?

Now you can see that the two parts of the positive value are added, and if you want the maximum value to be smaller, then the absolute value of the latter part must be = 0, so only the previous part is discussed.

f(x)=(x+a 2) squared, assuming that it is less than, (x+a 2) squared, you will know that the assumption is not true if you look at it algebraically.

The maximum value of this function is indefinite, so you'll want to discuss it first.

Because the function image opening up, so

When the axis of symmetry is x=-a 2<0(a>0), the maximum value is a+b, when -a 2>0(a<0), the maximum value is b-a, when -a 2=0(a=0), max=b+1 and b is uncertain, so... I can't solve it.

In fact, the easiest way is to combine numbers and shapes, you can draw an image of tanx, and you need to strictly prove that you use a concave and convex function.

(ab+a+b+1)(ab+ac+bc+c squared)>=(a+1)(b+1)(ab+2c*ab+c squared under the root number) = (a+1)(b+1)(root number ab +c) 2>=2 root number a*2 root number b*4c root number a*root number b = 16abc

Equal sign is established condition: a=b=c=1

ab+a+b+1=(a+1)(b+1)

ab+ac+bc+c^2=a(b+c)+c(b+c)=(a+c)(b+c)

a+1)(b+1)(a+c)(b+c)>=16 root number(a*b*ac*bc)=16abc

The parentheses are multiplied by the basic inequalities.

Is it necessary to defactor the factor? It's mean inequality! (ab+a+b+1)>=4 times ab*a*b*1 of the fourth root; (ab+ac+bc+c squared) > = 4 times ab*ac*bc*c square of the fourth root; The right side of the two inequalities ABC occurs four times, multiplied to form ABC, and the equal sign is taken at a=b=c=1 to be proved.

By the formula (a1+a2+..an) n> = open n root (a1a2....an) is the result, i.e., the arithmetic mean is greater than the geometric mean.

AB+A+B+1>=4x 4 times root (AABB)AB+AC+BC+C 2>=4x 4 times root (AABBCCCC) are multiplied to obtain the result.

For the last question, first of all, assuming that the restaurant owner has graduated from mathematics elementary school, then he should have a graduation certificate and the mathematics score on the graduation certificate should be greater than or equal to 60 points. You can launch the boss math number accuracy is greater than or equal to 99%, the correct rate of addition and subtraction operation is greater than or equal to 90%, the correct rate of multiplication and division operation is greater than or equal to 75%, the accuracy of comprehensive application operation is greater than or equal to 50%, and the ability to beat around the bush is greater than or equal to 50%. Then the probability that he will come out to sell steamed buns and make people eat them for nothing is less than 50%, and through the previous proof, people with a little brains can eat the boss's steamed buns for two cents, or even no money.

Therefore, the previous assumption is not valid, and it is concluded that the boss did not graduate from the elementary school of mathematics and was allowed to retake it.

1.Because 1 steamed bun has 4 corners, and 2 steamed buns should have 8 corners, it only costs 2 cents when buying a third one.

2.When you buy 5 steamed buns, you normally need 2 yuan, but according to 3 1 yuan, 2 yuan can buy 6 steamed buns, so you can eat one steamed bun without spending money.

Didn't understand the back - -

You can buy a steamed bun for two cents, buy two steamed buns for anise, and add two jiao to buy three for one yuan.

In other words, you can also buy a steamed bun in 2 corners!

You can eat a steamed bun without money:

Three for one yuan, six for two yuan, 4 corners for one to sell 5 can get two yuan plus a steamed bun.

So you don't need money to eat a steamed bun!

Let u=,v=, let the correlation vector of u be , and the correlation vector of v is , then dim(u)=p,dim(v)=q,u+v=,u+v is part of the correlation vector, assuming that is, then bs+1, bs+2 ,..BM can be expressed in it, then U w=, over dimu+v=n+s, dimu w=m-s, so it can be proved.