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What should I do if I can't do a math problem in high school, but I can understand the answer analysis, and I still can't do a similar problem?
The landlord's problem is actually quite typical, not a minor problem. Let me talk about some of my experiences in detail. There are many words. I hope the landlord can read it patiently. Ok, I'll give the landlord an example, and the landlord will understand.

For example, there is a question to prove 1+ 1/2? + 1/3? …+ 1/n? & lt2(n is a positive integer), the landlord can't do it after reading it, just look at the answer, which reads:

1+ 1/2? +…+ 1/n? & lt 1+ 1/( 1×2)+ 1/(2×3)+…+ 1/[n(n- 1)]

= 1+( 1- 1/2)+( 1/2- 1/3)+ 1/3-…+ 1/(n- 1)- 1/n

= 1+ 1- 1/2+ 1/2- 1/3+ 1/3-…+ 1/(n- 1)- 1/n

= 2- 1/n & lt; 2

Then the landlord took a closer look and made it clear. The first step is to make each denominator smaller and the numerical value larger; Step 2, disassemble each score in the form of 1/m(m+ 1); The third step is to cancel out a lot of addition, subtraction, multiplication and division in the middle, leaving 2- 1/n, so if it is less than 2, the proof is over. I understand every step.

Then I encountered a similar problem. Proof: 1+ 1/3? + 1/5? +…+ 1/n? & lt3/2(n is an odd number, n >;; 0), anyway, the above is understood, just like the above: the first step is to lower the denominator.

1+ 1/3? + 1/5? +…+ 1/n? & lt 1+ 1/(2×3)+ 1/(4×5)+…+ 1/(n- 1)n

Step 2, disassemble1+1/2-1/3+1/4 …+1/(n-1)-1/n.

Step three, offset a lot in the middle ... no! Can't offset it. What's going on here? I won't do it again after that. ...

What's the problem? The landlord's "understanding" or "understanding" is "a literal answer" I think the answer to a question has two parts, one is the "tangible part", that is, the answer is written on paper; The other is the "invisible part", that is, the idea, intention and source of the answer, and how to think of this problem-solving method from the topic. Literally, you only understand the first part, and the answer is step a- step b- step C (completion). Then you can understand what these three steps are, and you can understand the derivation of steps A to B and B to C, but the second part is difficult to understand. To understand the second part, we must understand many problems, such as "why do we adopt the a-b-c method", "how did we come up with the a-b-c method" and "why can't we adopt another method of a'-b'-c". Often do mechanical imitation, you only need to understand the first part, but you can do anything and change it. All similar questions can be done correctly, so you must understand the second part.

Let me talk about the "intangible part" of the answer at the beginning above. Ask a few questions first.

① Why should we narrow the denominator?

A: This is a common method to prove inequality, which is called "scale method".

② Why should the denominator be reduced according to this law?

A: Because of this, a score can be divided into two items: one positive and one negative.

(3) Why should it be split into two items?

A: What we want to prove is a summation form. We must find a variant to simplify the summation formula. The best way to simplify is the positive and negative cancellation of the intermediate term. At this time, you will find that the way to make the denominator smaller is not only to split the score, but also to cancel the middle term. After careful observation, it will be found that the key to offset is to make the end of the previous item and the beginning of the latter item the same number (for example,1(2× 3) and 1/(3×4) are all 3, which is the joint; If it is 1/(2×3) and 1/(4×5), it will not work.

Finally, we can sum up the "soul" of this kind of topic: simplify the denominator to the form of product, and the end of the previous term and the beginning of the latter term of the product are the same number, and then disassemble it to sum up the cancellation. Only by summing up this point can we understand the "intangible" part.

Knowing this, you can do similar problems. Can't imitate mechanically, put 1/3? It also becomes 1/(2×3), and then becomes 1/( 1×3), followed by 1/5? It becomes 1/(3×5) and so on, so that the difference between the two numbers on the denominator is 2, and it is docked.

1+ 1/3? + 1/5? +…+ 1/n? & lt 1+ 1/( 1×3)+ 1/(3×5)+…+ 1/[(n-2)n]

= 1+ 1/2[ 1- 1/3+ 1/3- 1/5+…+ 1/(n-2)- 1/n]

(Note that when the denominator differs by 2, it will be multiplied by 1/2 during disassembly. )

= 1+ 1/2- 1/(2n)= 3/2- 1/(2n)& lt; 3/2

This problem will be proved.

It is suggested that the landlord pay attention to two basic knowledge. Some seemingly problems on the topic are actually that the basic knowledge is not firmly mastered. To understand the answer thoroughly, it is often required that the knowledge points in your textbook should be solid, so that you can see what the intangible part of the answer is with knowledge sensitivity. Look at the answers and think more. Don't just ask "how did you get the answer from the first step to the second step", but also ask "how did you get the answer and what is the key to the success of this method".

Of course, in the end, when you are more proficient in mathematics, you will find that the "impossible" exercises in front of you can actually be done:

1+ 1/3? + 1/5? +…+ 1/n? & lt 1+ 1/(2×3)+ 1/(4×5)+…+ 1/(n- 1)n

= 1+ 1/2- 1/3+ 1/4+…+ 1/(n- 1)- 1/n

= 3/2+( 1/4- 1/3)+( 1/5- 1/4)+…- 1/n

You can see that the front is 3/2, followed by one-to-one combination (1/4- 1/3), (1/5- 1/4) ... All the results are negative, so it is generally 3/2 plus 1. Of course, this is obviously another way of thinking.