Mathematical induction is what type of problem to solve

To understand mathematical induction, it is highly recommended to play dominoes to get the most out of it!!

Proof steps: 1. Verify that n=n0 is true (n0 is the initial value of n) 2. assume that the original proposition is true when n k, and on this basis, prove that n k 1 is also true.

3. Conclude that for all natural numbers n n0, the original example is true.

Proof of the key points to note:

1. N=n0 must be verified, and this step is called the inductive basis (equivalent to toppling the first domino).

2. The key step is to assume that the original proposition is true when n k, and on this basis, it is proved that n k 1 is also true, this step is called the inductive hypothesis (the function is to prove that there is such a law between any two adjacent dominoes: if the former one falls, the latter must be able to fall), this step is also the most difficult.

3. In the process of proving that n k 1 is also true, it is necessary to use the conclusions obtained by the assumptions.

4. In the process of proving that n k 1 is also true, we should pay attention to two things: make up the form of the hypothesis, so as to make use of the conclusion of the hypothesis, and then make up the form of the proof result.

5. When n k 1, it is necessary to pay full attention to the difference with n k, and the addition or decrease of items.

It is usually used to solve proof and summarize inductive questions.

Verify: 1+2+3+......n=n(n+1)/2

When n=2, 1+2=2(1+2) 2=3, let n=m 1+2+3+......m=m(1+m) 2, then 1+2+3+......m+(m+1)=(m+1)+m(m+1) 2=(m+1)(1+m 2)=(m+1)(m+2) 2=(m+1)[(m+1)+1] 2 is certified.

This is a way to summarize relevant knowledge according to the type of questions in the college entrance examination. It helps to improve the speed and success rate of problem solving, and is very important for college entrance examination review. For example, the strategy for solving the inorganic block diagram problem is summarized

The chaos is based on characteristic reactions. Substances with special physical or chemical properties often have characteristic reactions or exhibit special phenomena in reactions. For example, the color reaction of the base mask is yellow, which is the characteristic of sodium; The gas that smells of rotten eggs is hydrogen sulfide; Turning blue in case of iodine is a characteristic of starch; The colorless gas that fades the magenta solution is sulfur dioxide; Nitric oxide turns reddish-brown when exposed to oxygen; The gas that turns the phenolphthalein test solution red is ammonia; White precipitation in the air from white - gray-green - reddish-brown is a characteristic reaction phenomenon of ferrous hydroxide to iron hydroxide and so on.

These characteristic reactions or phenomena can be used as a breakthrough in the solution of block diagram problems.

Solve according to the transformation relationship. Substances are inferred from the transformation relationship of the reaction, and the substance is confirmed by reading the diagram and thinking, and through screening and screening in the transformation relationship of common elements and compounds.

Take the information that recurs in the block diagram as a breakthrough.

Empathy and clever inference. Empathy means that when solving problems, you can adjust your way of thinking at any time according to the different problems and situations, such as changing conventional thinking to leaping thinking, changing common thinking to seeking different thinking, and changing positive thinking to reverse thinking, so as to achieve the effect of solving problems.

Use textbook knowledge and new information to make careful inferences.

Proof: (1).When n=1, left=1

Right = (1 4)·1·2 =1So left = right.

2）.When n=k, 1 +2 +3 +. k = (1 4) k (k+1) is true, then when n = k+1, the left = 1 +2 + 3 +. k²+（k+1)²

1/4)k²(k+1)²+k+1)²

k+1)²·1/4)k²+1】

Right = (1 4) (k+1) [k+1)+1] Left = right can be obtained by arrangement, so the original formula is established.

That's right.

Hypothetical methods can be used.

Compare k vs. k+1 again

The title is incorrect. 1^2+2^2+3^2+……n^2=(1/6)n(n+1)(2n+1)

1^3+2^3+3^3+……n^3=(1/4)n²(n+1)²

It can be proved by mathematical induction.

o point can't be the origin, right? If o is the origin, a regular triangle cannot be formed.

Proof: xn+1=xn 2+2 xn, and x1>=2.

When n=1, x2=x1 2+2 x1 2*open squared[(x1 2)*(2 x1)]=2>0;

Assuming that when n=k n*, xk>0 is true, then xk+1=xk 2+2 xk 2*open square[(xk 2)*(2 xk)]=2>0 is true;

Then when n=k+1 n*, xk+2=xk+1 2+2 xk+1 2>0 is true.

In summary, it can be concluded from mathematical induction that for any n n*, xn>0 is true.

This question is easy to prove with mathematical induction, as in the following sometimes, the answer system is very simple, just in case, the teacher usually strictly requires the steps, or write it is more safe, it is not very difficult, or to take up some space.

As for this obvious result, in my opinion, it is enough to mention it, and it proves that it should not be scored.

1x2x...1+r)+2x3x...2+r)+.n(n+1)..n+r)

n(n+1)(n+2)..n+r+1) r+2 is right.

It is important to note that the proof is true for all n, not for all r, n is the main variable and r is the argumentative variable, so don't be fooled by the form.

This formula is a universal form, when r=1, it is 1x2+2x3+3x4+.n(n+1)=n(n+1)(n+2)/3

When r=2, it is 1x2x3+2x3x4+3x4+3x4x5+.n(n+1)(n+2)=n(n+1)(n+2)(n+3) 4,r can take all positive integers, the formula is the identity of r, and has nothing to do with the value of r, just like x =x +1 -1 no matter what value x takes is true, we don't care about the specific value of r, but treat r as a fixed number (we don't care how many it is, we don't know how many it is, just like x above), and directly prove that the formula is true to r.

Mathematically inductive forensics 1x2x....1+r)+2x3x...2+r)+.n(n+1)..n+r) =n(n+1)(n+2)..n+r+1)/r+2

When n=1, there is left = 1x2x....1+r) =1x2x...1+r) (r+2) r+2 = right so the equation holds when n=1.

Suppose n=k holds the original formula, i.e., 1x2x....1+r)+2x3x...2+r)+.k(k+1)..k+r) =k(k+1)(k+2)..k+r+1)/r+2

When n=k+1, left =1x2x....1+r)+2x3x...2+r)+.

k(k+1)..k+r) +k+1)(k+2)..k+r) (k+r+1) = k(k+1)(k+2)..

k+r+1)/r+2 + k+1)(k+2)..k+r) (k+r+1) =

k+1)(k+2)..k+r) (k+r+1) *k/(r+2) +1] =(k+1)(k+2)..k+r) (k+r+1) *k+r+2) (r+2) = right.

That is, the equation is also true when n=k+1.

Therefore the formula is true for all n.

Through the proof, it can be seen that what is to be proved is true for all n, not for all r, n is the main variable, r is the paravariable, don't be confused by the form, the landlord's confusion lies in this.

Transform the equation.

n(n+1)(n+2)/3-n(n+1)(n-1)/3=n(n+1) sn-s(n-1)=an

n(n+1)(n+2)(n+3)/4-n(n+1)(n-1)/4=n(n+1)(n+2) sn-s(n-1)=an

It's the same equation.

At this point, you basically don't need any proof, it's all identities.

Often, they only know that left equals right and ignore that right equals left.

So this proof is equivalent to someone else giving (a+b) 2=a 2+2ab+b 2

If you write from right to left, that's a new conclusion.

That's why there are many places where this law isn't summed up, and people who understand it don't need to explain it.

It's great that you're thinking about this kind of problem, and if you have the opportunity, you'll have a copy of the method of the pending coefficients that can solve all ordinary summation problems.

n=1 is clearly true.

When n=k, 1x2x....1+r)+2x3x...2+r)+.k(k+1)..k+r) =k(k+1)(k+2)..k+r+1) r+2 is established.

Then n=k+1.

1x2x...1+r)+2x3x...2+r)+.k(k+1)..k+r)+(k+1)(k+2)..k+r+1)

k(k+1)(k+2)..k+r+1)/(r+2)+(k+1)(k+2)..k+r+1)

k+1)(k+2)..k+r+1)*[1+k/(r+2)](k+1)(k+2)..k+r+1)*(k+r+2) (r+2)The proposition holds for all n... The proposition is omitted after it is proven.

Let me answer it, can I answer the first question first?

1. Proof: Because the unit digit of the natural number multiple of 7 is 7 4 1 8 5 2 9 6 3 0, and the unit digit is 7 * 9 when the unit digit is 3

And because x,y >0 and is an integer. So, when x=9, 7x+10y>=63i.e. it is impossible to find x,y so that the equation 7x+10y=53 holds.

Certification. I'm going to answer it, and I want to say that in the second question, I think the proposition is wrong, and I'll prove him wrong.

2. The proposition is not true, prove: assuming that the proposition is true, it is proved by mathematical induction:

When n=54, there are x(1)=2 and y(1)=4 satisfied;

Let n=k hold, i.e. 7*x(k)+10*y(k)=k;

When n=k+1, 7*x(n)+10*y(n)=n=k+1

> 7*x(n)+10*y(n)=7*x(k)+10*y(k)+1

> 7*(x(n)-x(k))+10*(y(n)-y(k))=1

Solution: x(n)=x(k)+3

y(n)=y(k)-2

i.e. y decreases as n increases. Because n=54, y=4, then, when n=56, y=0, x=8It contradicts the requirement that x,y are positive integers.

That is, the proposition is not true.

In fact, we can conclude that the proposition is not true by enumerating the following, even if x,y are natural numbers:

n=55x=5,y=2

n=56x=8,y=0

n=57x=1,y=5

n=58x=4,y=3

n=59x=7,y=1

n=60x=10,y=-1

ps: What kind of mentality do those who think they think they are very good attest, do they carelessly prove that they are wrong and are pointed out by others?"Not powerful", do you want to take revenge on all the people who answer the questions and make the rest of the people useless? It's so awkward!

1. xy is a positive integer.

Let 7x+10y=53

Since the last digit of 10y can only be 0

Therefore, the end of 7x must be 3, and the end of x must be 9, and the minimum value of x is 9, so 7x=63 is not valid.

So 7x+10y=53 is not true.

2. Proof: The first step: 7x+10y=54, x=2, y=4 are trueThe second step: assuming that 7x1+10y1=n1 is true.

Then 7x2+10y2=n1+1 (3) is subtracted by the two equations:

7（x2-x1）+10（y2-y1）=1

It can be seen that when x2-x1=3

y2-y1=-2 can be established by equation (3).

1.The unit digit of 7x+10y is the same as the unit of 7y.

When the unit digit of x is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, the unit digit of 7x is 0, 7, 4, 1, 8, 5, 2, 9, 6, 3, which is exactly the same as all 10 digits.

If 7x+10y=53, then the unit digit of 7x is 3As can be seen from the above enumeration, the unit of x must be 9

Since x is a positive integer, x>=9So 7x+10y>=63, contradictory. The first question is proven.

2.The question is incorrect, because when n=56,60,63 there is no positive integer x,y. that satisfies the requirementsThe correct statement should be a non-negative integer.

Let 7x+10y=nWe perform an inductive proof of n.

When n=54, there is a positive integer solution (x,y)=(2,4);

When n=55, there is a positive integer solution (x,y)=(5,2);

When n=56, there is a positive integer solution (x,y)=(8,0);

When n=57, there is a positive integer solution (x,y)=(1,5);

When n=58, there is a positive integer solution (x,y)=(4,3);

When n=59, there is a positive integer solution (x,y)=(7,1);

When n=60, there is a positive integer solution (x,y)=(0,6);

When n=61, there is a positive integer solution (x,y)=(3,4);

When n=62, there is a positive integer solution(x,y)=(6,2);

When n=63, there is a positive integer solution (x,y)=(9,0)

Thus we can assume that proposition pairs 54, 55, 56 ,..n-1 holds, where n>=64

Since 54<=n-10, 7x+10y=n has a non-negative integer solution (a, b+1).

This proves that the proposition is true for n. According to the inductive method, the proposition is proven.