Some problems can be transformed into finding the sum of several numbers. To solve these problems, we need to judge whether to seek a sum of arithmetic progression. If it is arithmetic progression summation, we can use the summation formula of arithmetic sequence.

When solving the problem of the number of natural numbers, we should consider grouping the numbers in the questions appropriately according to the specific characteristics of the questions, and matching the numbers in each group reasonably, so as to solve the problem smoothly.

Second, be concise.

Liu Jun read a novel. The first day, he read 30 pages. From the next day, he read three more pages every day than the day before. On the day of 1 1, he read 60 pages, just finishing it. How many pages are there in this book?

According to the condition that "the number of pages he reads every day is three pages more than that of the previous day", we can know that the number of pages he reads every day is arranged according to a certain rule, that is, 30, 33, 36, ... 57, 60. How many pages does this book need, that is, the sum of the numbers in this column. The number of this column is arithmetic progression, the first item =30, the last item =60, and the number of items = 1 1. Therefore, the solution can be obtained quickly:

(30+60)× 1 1÷2 = 495(page)

Think about it: If "the first 1 1 day" is changed to "the last day", how to answer it?

Exercise 1:

1. Master Liu made a batch of parts. I made 30 yuan on the first day, so I made 2 yuan more every day than the day before. 15 made 48 yuan that day, just finished. How many parts are there in this batch?

2. Qian Hu reads a story book. On the first day, she read twenty pages. Since the next day, she has read five more pages every day than the day before. I just finished reading 50 pages on the last day. How many pages are there in this book?

Lily learned six English words on the first day, 1 words more than the day before, and 16 words on the last day. How many English words has Lili learned these days?

The keys of these locks are mixed up. How many times do you have to try to match each lock with your own key?

When thinking navigation opens the first lock, if unfortunately all 29 keys can't be opened, then the remaining one can be opened, that is, the first lock can be opened at most 29 times; Similarly, it takes up to 28 attempts to open the second lock, up to 27 attempts to open the third lock ... the 29th lock is opened, and the last one can be opened without trying. So try 29+28+27+…+2+1= (29+1) × 29 ÷ 2 = 435 at most.

Exercise 2:

1. The keys of 80 locks have been messed up. How many times should I try to match each lock with its own key?

The keys of some locks are mixed up. As we all know, it takes up to 28 attempts to match each lock with its own key. How many locks have a * * * key messed up?

3. There are 10 boxes and 44 shuttlecocks. Can you put 44 shuttlecocks in the box so that the number of shuttlecocks in each box is not equal?

There are 5 1 students in a class, and everyone shakes hands with others when they graduate. So how many times did * * * shake hands?

Thinking navigation assumes that 5 1 students are in a row. The first person shook hands with others in turn, one * * * shook hands with others 50 times, the second shook hands with others 49 times, and the third shook hands 48 times. By analogy, the 50th person shakes hands with the rest 1 time, so the sum of the times they shake hands is:

50+49+48+…+2+1= (50+1) × 50 ÷ 2 =1275 (times).

Exercise 3:

1. There is a table tennis match at school, and each player has to play a game with other players. If there are 2 1 people participating in the competition, how many competitions will there be for a * * *?

2. A class reunion, 43 students and 4 teachers, each student or teacher should shake hands with other students. How many times did you shake hands?

During the holiday, some students made an appointment to exchange telephone calls twice. They called 78 times and asked how many students had made an appointment to exchange phones.

Example 4 Find the sum of all the numbers of 99 continuous natural numbers, 1 ~ 99.

Thinking navigation must first make it clear that this problem is to find the sum of 99 consecutive natural numbers, not the sum of these 99 numbers. In order to solve the problem conveniently, we might as well count 0 first (without affecting our calculation of the sum) and calculate the sum of 100 from 0 to 99. After the numbers of 100 are paired, the sum of every two numbers is equal, that is, 9+9= 18, a * * *, and 100÷2=50 pairs. So the sum of all the 99 continuous natural numbers of 1 ~ 99 is 65438.

Exercise 4:

1. Find the sum of all numbers of 1 ~ 199 continuous natural numbers.

2. Find the sum of all the 999 consecutive natural numbers from 1 to 999.

3. Find the sum of all the 3000 consecutive natural numbers from 1 to 3000.

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Example 5 Find the sum of all numbers of 209 continuous natural numbers, 1 ~ 209.

For mental navigation, we might as well add all the numbers from 0 to 199, then add all the numbers from 200 to 209, and then combine them. The sum of all numbers from 0 to 199 is (1+9× 2 )× (200 ÷ 2) =1900, and the sum of all numbers from 200 to 209 is 2×10+/kloc-0. Therefore, the sum of all the 209 consecutive natural numbers 1 ~ 209 is 1900+65= 1965.

Exercise 5:

1. Find the sum of all numbers of 1 ~ 308 continuous natural numbers.

2. Find the sum of all continuous natural numbers from 1 to 2009.

3. Find the sum of all the continuous natural numbers from 2000 to 5000.