The denominator of the geometric progression is equal. The sum of an infinite geometric progression at. Problems to solve independently
NUMERIC SEQUENCES VI
§ l48. Sum infinitely decreasing geometric progression
Until now, when talking about sums, we have always assumed that the number of terms in these sums is finite (for example, 2, 15, 1000, etc.). But when solving some problems (especially higher mathematics) one has to deal with the sums of an infinite number of terms
S= a 1 + a 2 + ... + a n + ... . (1)
What are these amounts? A-priory the sum of an infinite number of terms a 1 , a 2 , ..., a n , ... is called the limit of the sum S n first P numbers when P -> ∞ :
S=S n = (a 1 + a 2 + ... + a n ). (2)
Limit (2), of course, may or may not exist. Accordingly, they say that the sum (1) exists or does not exist.
How can we find out whether sum (1) exists in each specific case? The general solution to this issue goes far beyond the scope of our program. However, there is one important special case, which we now have to consider. We will talk about summing the terms of an infinitely decreasing geometric progression.
Let a 1 , a 1 q , a 1 q 2, ... is an infinitely decreasing geometric progression. This means that | q |< 1. Сумма первых P terms of this progression is equal
From the basic theorems on the limits of variables (see § 136) we obtain:
But 1 = 1, a qn = 0. Therefore
So, the sum of an infinitely decreasing geometric progression is equal to the first term of this progression divided by one minus the denominator of this progression.
1) The sum of the geometric progression 1, 1/3, 1/9, 1/27, ... is equal to
and the sum of the geometric progression is 12; -6; 3; - 3 / 2 , ... equal
2) Convert a simple periodic fraction 0.454545 ... into an ordinary one.
To solve this problem, imagine this fraction as an infinite sum:
The right side of this equality is the sum of an infinitely decreasing geometric progression, the first term of which is equal to 45/100, and the denominator is 1/100. That's why
Using the described method, it can also be obtained general rule conversion of simple periodic fractions into ordinary ones (see Chapter II, § 38):
To convert a simple periodic fraction into an ordinary fraction, you need to do the following: put the period in the numerator decimal, and the denominator is a number consisting of nines taken as many times as there are digits in the period of the decimal fraction.
3) Convert the mixed periodic fraction 0.58333 .... into an ordinary fraction.
Let's imagine this fraction as an infinite sum:
On the right side of this equality, all terms, starting from 3/1000, form an infinitely decreasing geometric progression, the first term of which is equal to 3/1000, and the denominator is 1/10. That's why
Using the described method, a general rule for converting mixed periodic fractions into ordinary fractions can be obtained (see Chapter II, § 38). We deliberately do not present it here. There is no need to remember this cumbersome rule. It is much more useful to know that any mixed periodic fraction can be represented as the sum of an infinitely decreasing geometric progression and a certain number. And the formula
for the sum of an infinitely decreasing geometric progression, you must, of course, remember.
As an exercise, we suggest that you, in addition to the problems No. 995-1000 given below, once again turn to problem No. 301 § 38.
Exercises
995. What is called the sum of an infinitely decreasing geometric progression?
996. Find the sums of infinitely decreasing geometric progressions:
997. At what values X progression
is it infinitely decreasing? Find the sum of such a progression.
998. In an equilateral triangle with side A a new triangle is inscribed by connecting the midpoints of its sides; a new triangle is inscribed in this triangle in the same way, and so on ad infinitum.
a) the sum of the perimeters of all these triangles;
b) the sum of their areas.
999. Square with side A a new square is inscribed by connecting the midpoints of its sides; a square is inscribed in this square in the same way, and so on ad infinitum. Find the sum of the perimeters of all these squares and the sum of their areas.
1000. Compose an infinitely decreasing geometric progression such that its sum is equal to 25/4, and the sum of the squares of its terms is equal to 625/24.
Geometric progression is number sequence, the first term of which is non-zero, and each subsequent term is equal to the previous term multiplied by the same non-zero number. Geometric progression is denoted b1,b2,b3, …, bn, …
Properties of geometric progression
The ratio of any term of the geometric error to its previous term is equal to the same number, that is, b2/b1 = b3/b2 = b4/b3 = ... = bn/b(n-1) = b(n+1)/bn = … . This follows directly from the definition arithmetic progression. This number is called the denominator of a geometric progression. Usually the denominator of a geometric progression is denoted by the letter q.
One of the ways to specify a geometric progression is to specify its first term b1 and the denominator of the geometric error q. For example, b1=4, q=-2. These two conditions define the geometric progression 4, -8, 16, -32, ….
If q>0 (q is not equal to 1), then the progression is monotonous sequence. For example, the sequence, 2, 4,8,16,32, ... is a monotonically increasing sequence (b1=2, q=2).
If the denominator in the geometric error is q=1, then all terms of the geometric progression will be equal to each other. In such cases, the progression is said to be a constant sequence.
Formula for the nth term of the progression
In order for a number sequence (bn) to be a geometric progression, it is necessary that each of its members, starting from the second, be the geometric mean of neighboring members. That is, it is necessary to fulfill the following equation - (b(n+1))^2 = bn * b(n+2), for any n>0, where n belongs to the set natural numbers N.
The formula for the nth term of the geometric progression is:
bn=b1*q^(n-1), where n belongs to the set of natural numbers N.
Let's look at a simple example:
In geometric progression b1=6, q=3, n=8 find bn.
Let's use the formula for the nth term of a geometric progression.
Mathematics is whatpeople control nature and themselves.
Soviet mathematician, academician A.N. Kolmogorov
Geometric progression.
Along with problems on arithmetic progressions, problems related to the concept of geometric progression are also common in entrance examinations in mathematics. To successfully solve such problems, you need to know the properties of geometric progressions and have good skills in using them.
This article is devoted to the presentation of the basic properties of geometric progression. Examples of solving typical problems are also provided here., borrowed from the tasks of entrance examinations in mathematics.
Let us first note the basic properties of the geometric progression and recall the most important formulas and statements, associated with this concept.
Definition. A number sequence is called a geometric progression if each number, starting from the second, is equal to the previous one, multiplied by the same number. The number is called the denominator of a geometric progression.
For geometric progressionthe formulas are valid
, (1)
Where . Formula (1) is called the formula of the general term of a geometric progression, and formula (2) represents the main property of a geometric progression: each term of the progression coincides with the geometric mean of its neighboring terms and .
Note, that it is precisely because of this property that the progression in question is called “geometric”.
The above formulas (1) and (2) are generalized as follows:
, (3)
To calculate the amount first members of a geometric progressionformula applies
If we denote , then
Where . Since , formula (6) is a generalization of formula (5).
In the case when and geometric progressionis infinitely decreasing. To calculate the amountof all terms of an infinitely decreasing geometric progression, the formula is used
. (7)
For example , using formula (7) we can show, What
Where . These equalities are obtained from formula (7) under the condition that , (first equality) and , (second equality).
Theorem. If , then
Proof. If , then
The theorem has been proven.
Let's move on to consider examples of solving problems on the topic “Geometric progression”.
Example 1. Given: , and . Find .
Solution. If we apply formula (5), then
Answer: .
Example 2. Let it be. Find .
Solution. Since and , we use formulas (5), (6) and obtain a system of equations
If the second equation of system (9) is divided by the first, then or . It follows from this that . Let's consider two cases.
1. If, then from the first equation of system (9) we have.
2. If , then .
Example 3. Let , and . Find .
Solution. From formula (2) it follows that or . Since , then or .
By condition . However, therefore. Since and then here we have a system of equations
If the second equation of the system is divided by the first, then or .
Since, the equation has a unique suitable root. In this case, it follows from the first equation of the system.
Taking into account formula (7), we obtain.
Answer: .
Example 4. Given: and . Find .
Solution. Since, then.
Since , then or
According to formula (2) we have . In this regard, from equality (10) we obtain or .
However, by condition, therefore.
Example 5. It is known that . Find .
Solution. According to the theorem, we have two equalities
Since , then or . Because , then .
Answer: .
Example 6. Given: and . Find .
Solution. Taking into account formula (5), we obtain
Since, then. Since , and , then .
Example 7. Let it be. Find .
Solution. According to formula (1) we can write
Therefore, we have or . It is known that and , therefore and .
Answer: .
Example 8. Find the denominator of an infinite decreasing geometric progression if
And .
Solution. From formula (7) it follows And . From here and from the conditions of the problem we obtain a system of equations
If the first equation of the system is squared, and then divide the resulting equation by the second equation, then we get
Or .
Answer: .
Example 9. Find all values for which the sequence , , is a geometric progression.
Solution. Let , and . According to formula (2), which defines the main property of a geometric progression, we can write or .
From here we get the quadratic equation, whose roots are And .
Let's check: if, then , and ;
if , then , and . In the first case we have
and , and in the second – and .
Answer: , .Example 10.
, (11)
Solve the equation
where and .
From formula (7) it follows, What Solution. The left side of equation (11) is the sum of an infinite decreasing geometric progression, in which and , subject to: and .. In this regard, equation (11) takes the form or . Suitable root quadratic equation
Answer: .
is Example 11. P consistency positive numbers forms an arithmetic progression , A– geometric progression
Solution., what does it have to do with . Find . Because arithmetic sequence , That(the main property of arithmetic progression). Because the , then or . This implies ,that the geometric progression has the form. According to formula (2)
, then we write down that . Since and , then. In this case, the expression takes the form or . By condition ,so from Eq. we obtain a unique solution to the problem under consideration
Answer: .
, i.e. . Example 12.
. (12)
Solution. Calculate Sum
Multiply both sides of equality (12) by 5 and get arithmetic sequence
If we subtract (12) from the resulting expression
or .
Answer: .
To calculate, we substitute the values \u200b\u200binto formula (7) and get . Since, then. The examples of problem solving given here will be useful to applicants in preparing for entrance examinations, . For a deeper study of problem solving methods, related to geometric progression can be used teaching aids
from the list of recommended literature.
1. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Mir and Education, 2013. – 608 p. 2. Suprun V.P. Mathematics for high school students: additional sections school curriculum. – M.: Lenand / URSS
, 2014. – 216 p. 3. Medynsky M.M. Full course elementary mathematics in problems and exercises. Book 2: Number Sequences and Progressions. – M.: Editus
, 2015. – 208 p.
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This number is called the denominator of a geometric progression, i.e. each term differs from the previous one by q times. (We will assume that q ≠ 1, otherwise everything is too trivial). It's not hard to see that general formula nth term of the geometric progression b n = b 1 q n – 1 ; terms with numbers b n and b m differ by q n – m times.Already in Ancient Egypt knew not only arithmetic, but also geometric progression. Here, for example, is a problem from the Rhind papyrus: “Seven faces have seven cats; Each cat eats seven mice, each mouse eats seven ears of corn, and each ear of barley can grow seven measures of barley. How large are the numbers in this series and their sum?
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Rice. 1. Ancient Egyptian geometric progression problem |
This task was repeated many times with different variations among other peoples at other times. For example, in written in the 13th century. “The Book of the Abacus” by Leonardo of Pisa (Fibonacci) has a problem in which 7 old women appear on their way to Rome (obviously pilgrims), each of whom has 7 mules, each of which has 7 bags, each of which contains 7 loaves , each of which has 7 knives, each of which has 7 sheaths. The problem asks how many objects there are.
The sum of the first n terms of the geometric progression S n = b 1 (q n – 1) / (q – 1) . This formula can be proven, for example, like this: S n = b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n – 1.
Add the number b 1 q n to S n and get:
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From here S n (q – 1) = b 1 (q n – 1), and we get the necessary formula.
Already on one of the clay tablets of Ancient Babylon, dating back to the 6th century. BC e., contains the sum 1 + 2 + 2 2 + 2 3 + ... + 2 9 = 2 10 – 1. True, as in a number of other cases, we do not know how this fact was known to the Babylonians.
The rapid increase in geometric progression in a number of cultures, in particular in Indian, is repeatedly used as a visual symbol of the vastness of the universe. IN famous legend On the advent of chess, the ruler gives its inventor the opportunity to choose the reward himself, and he asks for the number of wheat grains that would be obtained if one were placed on the first square of the chessboard, two on the second, four on the third, eight on the fourth, etc. ., each time the number doubles. Vladyka thought that at most we were talking about a few bags, but he miscalculated. It is easy to see that for all 64 squares of the chessboard the inventor would have to receive (2 64 - 1) grains, which is expressed as a 20-digit number; even if the entire surface of the Earth was sown, it would take at least 8 years to collect the required amount of grains. This legend is sometimes interpreted as indicating the virtually unlimited possibilities hidden in the game of chess.
It is easy to see that this number is really 20-digit:
2 64 = 2 4 ∙ (2 10) 6 = 16 ∙ 1024 6 ≈ 16 ∙ 1000 6 = 1.6∙10 19 (a more accurate calculation gives 1.84∙10 19). But I wonder if you can find out what digit this number ends with?
A geometric progression can be increasing if the denominator is greater than 1, or decreasing if it is less than one. In the latter case, the number q n for sufficiently large n can become arbitrarily small. While the increasing geometric progression increases unexpectedly quickly, the decreasing geometric progression decreases just as quickly.
The larger n, the weaker the number q n differs from zero, and the closer the sum of n terms of the geometric progression S n = b 1 (1 – q n) / (1 – q) to the number S = b 1 / (1 – q). (For example, F. Viet reasoned this way). The number S is called the sum of an infinitely decreasing geometric progression. However, for many centuries the question of what is the meaning of summing the ENTIRE geometric progression, with its infinite number of terms, was not clear enough to mathematicians.
A decreasing geometric progression can be seen, for example, in Zeno’s aporias “Half Division” and “Achilles and the Tortoise.” In the first case, it is clearly shown that the entire road (assuming length 1) is the sum of an infinite number of segments 1/2, 1/4, 1/8, etc. This is, of course, the case from the point of view of ideas about a finite sum infinite geometric progression. And yet - how can this be?
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Rice. 2. Progression with a coefficient of 1/2 |
In the aporia about Achilles, the situation is a little more complicated, because here the denominator of the progression is not 1/2, but some other number. Let, for example, Achilles run with speed v, the tortoise moves with speed u, and the initial distance between them is l. Achilles will cover this distance in time l/v, and during this time the turtle will move a distance lu/v. When Achilles runs this segment, the distance between him and the turtle will become equal to l (u /v) 2, etc. It turns out that catching up with the turtle means finding the sum of an infinitely decreasing geometric progression with the first term l and the denominator u /v. This sum - the segment that Achilles will eventually run to the meeting place with the turtle - is equal to l / (1 – u /v) = lv / (v – u). But, again, how to interpret this result and why it makes any sense at all was not very clear for a long time.
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Rice. 3. Geometric progression with a coefficient of 2/3 |
Archimedes used the sum of a geometric progression to determine the area of a parabola segment. Let this segment of the parabola be delimited by the chord AB and let the tangent at point D of the parabola be parallel to AB. Let C be the midpoint of AB, E the midpoint of AC, F the midpoint of CB. Let's draw lines parallel to DC through points A, E, F, B; Let the tangent drawn at point D intersect these lines at points K, L, M, N. Let's also draw segments AD and DB. Let the line EL intersect the line AD at point G, and the parabola at point H; line FM intersects line DB at point Q, and the parabola at point R. According to general theory conic sections, DC – diameter of the parabola (that is, a segment parallel to its axis); it and the tangent at point D can serve as coordinate axes x and y, in which the equation of the parabola is written as y 2 = 2px (x is the distance from D to any point of a given diameter, y is the length of a segment parallel to a given tangent from this point of diameter to some point on the parabola itself).
By virtue of the parabola equation, DL 2 = 2 ∙ p ∙ LH, DK 2 = 2 ∙ p ∙ KA, and since DK = 2DL, then KA = 4LH. Because KA = 2LG, LH = HG. The area of segment ADB of a parabola is equal to the area of triangle ΔADB and the areas of segments AHD and DRB combined. In turn, the area of the segment AHD is similarly equal to the area of the triangle AHD and the remaining segments AH and HD, with each of which you can perform the same operation - split into a triangle (Δ) and the two remaining segments (), etc.:
The area of the triangle ΔAHD is equal to half the area of the triangle ΔALD (they have a common base AD, and the heights differ by 2 times), which, in turn, is equal to half the area of the triangle ΔAKD, and therefore half the area of the triangle ΔACD. Thus, the area of the triangle ΔAHD is equal to a quarter of the area of the triangle ΔACD. Likewise, the area of triangle ΔDRB is equal to one-quarter of the area of triangle ΔDFB. So, the areas of the triangles ΔAHD and ΔDRB, taken together, are equal to a quarter of the area of the triangle ΔADB. Repeating this operation when applied to segments AH, HD, DR and RB will select triangles from them, the area of which, taken together, will be 4 times less than the area of triangles ΔAHD and ΔDRB, taken together, and therefore 16 times less, than the area of the triangle ΔADB. And so on:
Thus, Archimedes proved that “every segment contained between a straight line and a parabola constitutes four-thirds of a triangle having the same base and equal height.”
A geometric progression is a numerical sequence, the first term of which is non-zero, and each subsequent term is equal to the previous term multiplied by the same non-zero number.
Geometric progression is denoted b1,b2,b3, …, bn, … .
The ratio of any term of the geometric error to its previous term is equal to the same number, that is, b2/b1 = b3/b2 = b4/b3 = ... = bn/b(n-1) = b(n+1)/bn = … . This follows directly from the definition of an arithmetic progression. This number is called the denominator of a geometric progression. Usually the denominator of a geometric progression is denoted by the letter q.
Monotonous and constant sequence
One of the ways to specify a geometric progression is to specify its first term b1 and the denominator of the geometric error q. For example, b1=4, q=-2. These two conditions define the geometric progression 4, -8, 16, -32, ….
If q>0 (q is not equal to 1), then the progression is monotonous sequence. For example, the sequence, 2, 4,8,16,32, ... is a monotonically increasing sequence (b1=2, q=2).
If the denominator in the geometric error is q=1, then all terms of the geometric progression will be equal to each other. In such cases they say that progression is constant sequence.
Formula for the nth term of a geometric progression
In order for a number sequence (bn) to be a geometric progression, it is necessary that each of its members, starting from the second, be the geometric mean of neighboring members. That is, it is necessary to fulfill the following equation
(b(n+1))^2 = bn * b(n+2), for any n>0, where n belongs to the set of natural numbers N.
The formula for the nth term of the geometric progression is:
bn=b1*q^(n-1),
where n belongs to the set of natural numbers N.
Formula for the sum of the first n terms of a geometric progression
The formula for the sum of the first n terms of a geometric progression has the form:
Sn = (bn*q - b1)/(q-1), where q is not equal to 1.
Let's look at a simple example:
In geometric progression b1=6, q=3, n=8 find Sn.
To find S8, we use the formula for the sum of the first n terms of a geometric progression.
S8= (6*(3^8 -1))/(3-1) = 19,680.