1+2+3+4+…+99+ 100=?
After the teacher finished the problem, the whole class was immersed in calculation, but Gao Xiao Jones quickly figured out that the answer was equal to 5050. Why does Gauss calculate quickly and accurately? It turns out that little Gauss found through careful observation that:
1+ 100=2+99=3+98=…=49+52=50+5 1。
1 ~ 100 can be divisible by 50 logarithms, and the sum of each logarithms is equal. So, little Gauss cleverly calculated this problem as
( 1+ 100)× 100÷2=5050。
This summation method used by Little Gauss is really ingenious, simple and quick, and widely applicable to arithmetic progression's summation problem.
Arranging several numbers in a row is called a series, and each number in the series is called an item, in which the first item is called the first item and the last item is called the last item. The difference between the last term and the previous term of a series is equal to arithmetic progression, and the difference between the last term and the previous term is called tolerance series. For example:
( 1) 1,2,3,4,5,…, 100;
(2) 1,3,5,7,9,…,99;
(3)8, 15,22,29,36,…,7 1。
Where (1) is the arithmetic progression with the first term of 1, the last term of 100 and the tolerance of 1; (2) arithmetic progression whose first term is 1 and the last term is 99 with a tolerance of 2; (3) It is a arithmetic progression with the first term of 8, the last term of 7 1 and the tolerance of 7.
Arithmetic progression's summation formula is obtained by Gauss's ingenious calculation method:
Sum = (first item+last item) × number of items ÷2.
Example11+2+3+…+1999 =?
Analysis and solution: this series of addends 1, 2, 3, …, 1999 are arithmetic progression, the first term is 1, and the last term is 1999. * * There is a number of 1999. It can be obtained from the summation formula of arithmetic sequence
The original formula = (1+1999) ×1999 ÷ 2 =199000.
Note: Before using the arithmetic sequence summation formula, it is necessary to judge whether each addend in the topic constitutes a arithmetic progression.
Example 211+12+13+…+31=?
Analysis and solution: the addend of this series is 1 1, 12, 13, …, 3 1 is arithmetic progression, the first term is 1 1, and the last term is 3/kloc.
The original formula = (11+31) × 21÷ 2 = 441.
When using the sum formula of arithmetic sequence, sometimes the number of terms is not clear at a glance, so it is necessary to find the number of terms first. According to the relationship between the first term, the last term and the tolerance, we can get
Number of items = (last item-first item) ÷ tolerance+1,
The last project = the first project+tolerance × (project number-1).
Example 3 3+7+ 1 1+…+99 =?
Analysis and solution: 3,7, 1 1, …, 99 is arithmetic progression with a tolerance of 4,
Project number = (99-3) ÷ 4+ 1 = 25,
The original formula = (3+99) × 25 ÷ 2 = 1275.
Example 4 Find the sum of the first 40 items in arithmetic progression, the first item is 25, and the tolerance is 3.
Solution: The last term = 25+3× (40- 1) = 142,
And = (25+ 142) × 40 ÷ 2 = 3340.
Arithmetic progression's summation formula and the formula for finding the number of terms and the final term can be used to solve various problems related to arithmetic progression's summation.
Example 5 In the figure below, the area of each smallest equilateral triangle is 12cm2, and the side length is 1 matchstick. Q: (1) What is the area of the largest triangle in square centimeters? (2) How many matchsticks does the whole figure consist of?
Analysis: The largest triangle * * * has 8 layers. When swinging from top to bottom, the number of small triangles in each layer and the number of matches used are as follows:
As can be seen from the above table, the number of small triangles in each layer is arithmetic progression, and the matching number in each layer is also arithmetic progression.
Solution: (1) The maximum triangle area is
( 1+3+5+…+ 15)× 12
=[( 1+ 15)×8÷2]× 12
= 768 (square centimeter).
(2) The number of matchsticks is
3+6+9+…+24
= (3+24) × 8 ÷ 2 = 108 (root).
A: The area of the largest triangle is 768cm2, and the whole figure consists of 108 matches.
There are three ping-pong balls in the box. A magician took a ball out of the box for the first time, turned it into three balls, and then put it back in the box. The second time, I took out two balls from the box, changed each ball into three balls, and then put them back in the box ... The tenth time, I took out ten balls from the box, changed each ball into three balls and put them back in the box. How many ping-pong balls are there in the box at the moment?
Analysis and solution: one ball becomes three balls, but actually there are two more balls. Two more balls for the first time, two more balls for the second time ... and two more balls for the tenth time. So after eating it ten times, it's much more.
2× 1+2×2+…+2× 10
=2×( 1+2+…+ 10)
= 2× 55 = 1 10 (only).
Plus the original three balls, there are balls in the box * * *110+3 =113 (only).
The comprehensive formula is:
(3- 1)×( 1+2+…+ 10)+3
= 2× [(1+10 )×10 ÷ 2]+3 =113 (only).
Exercise 3
1. Calculate the following question:
( 1)2+4+6+…+200;
(2) 17+ 19+2 1+…+39;
(3)5+8+ 1 1+ 14+…+50;
(4)3+ 10+ 17+24+…+ 10 1。
2. Find the sum of arithmetic progression with the first term of 5, the last term of 93 and the tolerance of 4.
3. Find the sum of the top 30 items in arithmetic progression, the first item is 13, and the tolerance is 5.
The clock strikes the hour, and the number of hours is equal to the number of hours, and it strikes the hour every half hour. Q: How many times does the clock strike day and night?
5. Find the sum of all numbers in 100 divided by 3 and 2.
6. Of all the two digits, how many digits * * * are greater than single digits?