How to find the denominator of a geometric progression. Infinitely decreasing geometric progression

NUMERICAL SEQUENCES VI

§ l48. The sum of an infinitely decreasing geometric progression

Until now, speaking of 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, the sum (1) is said to exist or not to exist.

How to find out whether the sum (1) exists in each particular case? A general solution to this question goes far beyond the scope of our program. However, there is one important special case that we have to consider now. We will talk about the summation of the terms of an infinitely decreasing geometric progression.

Let be a 1 , a 1 q , a 1 q 2 , ... is an infinitely decreasing geometric progression. This means that | q |< 1. Сумма первых P members of this progression is equal to

From the basic theorems on the limits of variables (see § 136) we obtain:

But 1 = 1, a q n = 0. Therefore

So, the sum of an infinitely decreasing geometric progression is equal to the first term of this progress divided by one minus the denominator of this progression.

1) The sum of the geometric progression 1, 1/3, 1/9, 1/27, ... is

and the sum of a geometric progression is 12; -6; 3; - 3 / 2 , ... equals

2) A simple periodic fraction 0.454545 ... turn into an ordinary one.

To solve this problem, we represent 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 45/100, and the denominator is 1/100. So

In the manner described, the general rule for converting simple periodic fractions into ordinary fractions can also be obtained (see Chapter II, § 38):

To convert a simple periodic fraction into an ordinary one, you need to proceed as follows: put the period of the decimal fraction in the numerator, and in the denominator - a number consisting of nines taken as many times as there are digits in the period of the decimal fraction.

3) Mixed periodic fraction 0.58333 .... turn into an ordinary fraction.

Let's represent 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 3/1000, and the denominator is 1/10. So

In the manner described, the general rule for the conversion of mixed periodic fractions into ordinary fractions can also be obtained (see Chapter II, § 38). We deliberately do not include it here. There is no need to memorize 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 some number. And the formula

for the sum of an infinitely decreasing geometric progression, one 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 sums of infinitely decreasing geometric progressions:

997. For what values X progression

is infinitely decreasing? Find the sum of such a progression.

998. In an equilateral triangle with a 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. In a square with a 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. Make 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.

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 is easy to see that the general formula of the nth member of the geometric progression is 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, they knew not only arithmetic, but also geometric progression. Here, for example, is a task from the Rhind papyrus: “Seven faces have seven cats; each cat eats seven mice, each mouse eats seven ears of corn, each ear can grow seven measures of barley. How large are the numbers in this series and their sum?

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 XIII 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 which has 7 mules, each of which has 7 bags, each of which contains 7 loaves , each of which has 7 knives, each of which is in 7 sheaths. The problem asks how many items there are.

The sum of the first n members of the geometric progression S n = b 1 (q n - 1) / (q - 1) . This formula can be proved, for example, as follows: S n \u003d b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n - 1.

Let's add the number b 1 q n to S n and get:

S n + b 1 q n = b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n – 1 + b 1 q n = b 1 + (b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n –1) q = b 1 + S n q .

Hence 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 VI 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 where this fact was known to the Babylonians.

The rapid growth of geometric progression in a number of cultures, in particular, in India, is repeatedly used as a visual symbol of the immensity of the universe. In the well-known legend about the appearance of chess, the ruler gives their inventor the opportunity to choose a reward himself, and he asks for such a number of wheat grains as will be obtained if one is placed on the first cell of the chessboard, two on the second, four on the third, eight on the fourth, and etc., each time the number is doubled. Vladyka thought that it was, at the most, a few sacks, but he miscalculated. It is easy to see that for all 64 squares of the chessboard the inventor should have received (2 64 - 1) grain, 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 number of grains. This legend is sometimes interpreted as a reference to the almost unlimited possibilities hidden in the game of chess.

The fact that this number is really 20-digit is easy to see:

2 64 \u003d 2 4 ∙ (2 10) 6 \u003d 16 1024 6 ≈ 16 1000 6 \u003d 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 is increasing if the denominator is greater than 1 in absolute value, or decreasing if it is less than one. In the latter case, the number q n can become arbitrarily small for sufficiently large n. While an increasing exponential increases unexpectedly fast, a decreasing exponential decreases just as quickly.

The larger n, the weaker the number q n differs from zero, and the closer the sum of n members of the geometric progression S n \u003d b 1 (1 - q n) / (1 - q) to the number S \u003d b 1 / (1 - q) . (So ​​reasoned, for example, F. Viet). 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 the summation of the ALL 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 "Biting" and "Achilles and the tortoise". In the first case, it is clearly shown that the entire road (assume length 1) is the sum of an infinite number of segments 1/2, 1/4, 1/8, etc. This, of course, is how it is from the point of view of ideas about the finite sum infinite geometric progression. And yet - how can this be?

Rice. 2. Progression with a factor of 1/2

In the aporia about Achilles, the situation is a little more complicated, because here the denominator of the progression is not equal to 1/2, but to some other number. Let, for example, Achilles run at speed v, the tortoise moves at speed u, and the initial distance between them is l. Achilles will run this distance in the time l / v , the tortoise will move a distance lu / v during this time. When Achilles runs through 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 point with the turtle - is equal to l / (1 - u / v) = lv / (v - u) . But, again, how this result should be interpreted and why it makes any sense at all, was not very clear for a long time.

Rice. 3. Geometric progression with coefficient 2/3

The sum of a geometric progression was used by Archimedes when determining the area of ​​a segment of a parabola. Let the given segment of the parabola be delimited by the chord AB and let the tangent at the 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 . Draw lines parallel to DC through points A , E , F , B ; let the tangent drawn at point D , these lines intersect at points K , L , M , N . Let's also draw segments AD and DB. Let the line EL intersect the line AD at the point G, and the parabola at the point H; line FM intersects line DB at point Q, and the parabola at point R. According to the general theory of conic sections, DC is the diameter of a 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 parabola equation is written as y 2 \u003d 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 . Since KA = 2LG , LH = HG . The area of ​​the segment ADB of the parabola is equal to the area of ​​the triangle ΔADB and the areas of the segments AHD and DRB combined. In turn, the area of ​​the AHD segment is similarly equal to the area of ​​the triangle AHD and the remaining segments AH and HD, with each of which the same operation can be performed - 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 ​​triangle ΔAHD is equal to a quarter of the area of ​​triangle ΔACD. Likewise, the area of ​​triangle ΔDRB is equal to a quarter of the area of ​​triangle ΔDFB. So, the areas of triangles ∆AHD and ∆DRB, taken together, are equal to a quarter of the area of ​​triangle ∆ADB. Repeating this operation as applied to the segments AH , HD , DR and RB will also 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 . Etc:

Thus, Archimedes proved that "every segment enclosed between a straight line and a parabola is four-thirds of a triangle, having with it the same base and equal height."

>>Math: Geometric progression

For the convenience of the reader, this section follows exactly the same plan as we followed in the previous section.

1. Basic concepts.

Definition. A numerical sequence, all members of which are different from 0 and each member of which, starting from the second, is obtained from the previous member by multiplying it by the same number is called a geometric progression. In this case, the number 5 is called the denominator of a geometric progression.

Thus, a geometric progression is a numerical sequence (b n) given recursively by the relations

Is it possible, by looking at a number sequence, to determine whether it is a geometric progression? Can. If you are convinced that the ratio of any member of the sequence to the previous member is constant, then you have a geometric progression.
Example 1

1, 3, 9, 27, 81,... .
b 1 = 1, q = 3.

Example 2

This is a geometric progression that
Example 3

This is a geometric progression that
Example 4

8, 8, 8, 8, 8, 8,....

This is a geometric progression where b 1 - 8, q = 1.

Note that this sequence is also an arithmetic progression (see Example 3 from § 15).

Example 5

2,-2,2,-2,2,-2.....

This is a geometric progression, in which b 1 \u003d 2, q \u003d -1.

Obviously, a geometric progression is an increasing sequence if b 1 > 0, q > 1 (see Example 1), and a decreasing sequence if b 1 > 0, 0< q < 1 (см. пример 2).

To indicate that the sequence (b n) is a geometric progression, the following notation is sometimes convenient:

The icon replaces the phrase "geometric progression".
We note one curious and at the same time quite obvious property of a geometric progression:
If the sequence is a geometric progression, then the sequence of squares, i.e. is a geometric progression.
In the second geometric progression, the first term is equal to a equal to q 2.
If we discard all the terms following b n exponentially, then we get a finite geometric progression
In the following paragraphs of this section, we will consider the most important properties of a geometric progression.

2. Formula of the n-th term of a geometric progression.

Consider a geometric progression denominator q. We have:

It is not difficult to guess that for any number n the equality

This is the formula for the nth term of a geometric progression.

Comment.

If you have read the important remark from the previous paragraph and understood it, then try to prove formula (1) by mathematical induction, just as it was done for the formula of the nth term of an arithmetic progression.

Let's rewrite the formula of the nth term of the geometric progression

and introduce the notation: We get y \u003d mq 2, or, in more detail,
The argument x is contained in the exponent, so such a function is called an exponential function. This means that a geometric progression can be considered as an exponential function given on the set N of natural numbers. On fig. 96a shows a graph of the function of Fig. 966 - function graph In both cases, we have isolated points (with abscissas x = 1, x = 2, x = 3, etc.) lying on some curve (both figures show the same curve, only differently located and depicted in different scales). This curve is called the exponent. More about the exponential function and its graph will be discussed in the 11th grade algebra course.

Let's return to examples 1-5 from the previous paragraph.

1) 1, 3, 9, 27, 81,... . This is a geometric progression, in which b 1 \u003d 1, q \u003d 3. Let's make a formula for the nth term
2) This is a geometric progression, in which Let's formulate the n-th term

This is a geometric progression that Compose the formula for the nth term
4) 8, 8, 8, ..., 8, ... . This is a geometric progression, in which b 1 \u003d 8, q \u003d 1. Let's make a formula for the nth term
5) 2, -2, 2, -2, 2, -2,.... This is a geometric progression, in which b 1 = 2, q = -1. Compose the formula for the nth term

Example 6

Given a geometric progression

In all cases, the solution is based on the formula of the nth member of a geometric progression

a) Putting n = 6 in the formula of the nth term of the geometric progression, we get

b) We have

Since 512 \u003d 2 9, we get n - 1 \u003d 9, n \u003d 10.

d) We have

Example 7

The difference between the seventh and fifth members of the geometric progression is 48, the sum of the fifth and sixth members of the progression is also 48. Find the twelfth member of this progression.

First stage. Drawing up a mathematical model.

The conditions of the task can be briefly written as follows:

Using the formula of the n-th member of a geometric progression, we get:
Then the second condition of the problem (b 7 - b 5 = 48) can be written as

The third condition of the problem (b 5 +b 6 = 48) can be written as

As a result, we obtain a system of two equations with two variables b 1 and q:

which, in combination with condition 1) written above, is the mathematical model of the problem.

Second phase.

Working with the compiled model. Equating the left parts of both equations of the system, we get:

(we have divided both sides of the equation into the expression b 1 q 4 , which is different from zero).

From the equation q 2 - q - 2 = 0 we find q 1 = 2, q 2 = -1. Substituting the value q = 2 into the second equation of the system, we obtain
Substituting the value q = -1 into the second equation of the system, we get b 1 1 0 = 48; this equation has no solutions.

So, b 1 \u003d 1, q \u003d 2 - this pair is the solution to the compiled system of equations.

Now we can write down the geometric progression in question: 1, 2, 4, 8, 16, 32, ... .

Third stage.

The answer to the problem question. It is required to calculate b 12 . We have

Answer: b 12 = 2048.

3. The formula for the sum of members of a finite geometric progression.

Let there be a finite geometric progression

Denote by S n the sum of its terms, i.e.

Let's derive a formula for finding this sum.

Let's start with the simplest case, when q = 1. Then the geometric progression b 1 ,b 2 , b 3 ,..., bn consists of n numbers equal to b 1 , i.e. the progression is b 1 , b 2 , b 3 , ..., b 4 . The sum of these numbers is nb 1 .

Let now q = 1 To find S n we use an artificial method: let's perform some transformations of the expression S n q. We have:

Performing transformations, we, firstly, used the definition of a geometric progression, according to which (see the third line of reasoning); secondly, they added and subtracted why the meaning of the expression, of course, did not change (see the fourth line of reasoning); thirdly, we used the formula of the n-th member of a geometric progression:

From formula (1) we find:

This is the formula for the sum of n members of a geometric progression (for the case when q = 1).

Example 8

Given a finite geometric progression

a) the sum of the members of the progression; b) the sum of the squares of its terms.

b) Above (see p. 132) we have already noted that if all members of a geometric progression are squared, then a geometric progression with the first member b 2 and the denominator q 2 will be obtained. Then the sum of the six terms of the new progression will be calculated by

Example 9

Find the 8th term of a geometric progression for which

In fact, we have proved the following theorem.

A numerical sequence is a geometric progression if and only if the square of each of its terms, except for the first one (and the last one, in the case of a finite sequence), is equal to the product of the previous and subsequent terms (a characteristic property of a geometric progression).

Consider now the question of summation of an infinite geometric progression. Let us call the partial sum of a given infinite progression the sum of its first terms. Denote the partial sum by the symbol

For every infinite progression

one can compose a (also infinite) sequence of its partial sums

Let a sequence with unlimited increase have a limit

In this case, the number S, i.e., the limit of partial sums of the progression, is called the sum of an infinite progression. We will prove that an infinite decreasing geometric progression always has a sum, and derive a formula for this sum (we can also show that for an infinite progression has no sum, does not exist).

We write the expression for the partial sum as the sum of the members of the progression according to formula (91.1) and consider the limit of the partial sum at

From the theorem of item 89 it is known that for a decreasing progression ; therefore, applying the difference limit theorem, we find

(the rule is also used here: the constant factor is taken out of the sign of the limit). The existence is proved, and at the same time the formula for the sum of an infinitely decreasing geometric progression is obtained:

Equality (92.1) can also be written as

Here it may seem paradoxical that a well-defined finite value is assigned to the sum of an infinite set of terms.

A clear illustration can be given to explain this situation. Consider a square with a side equal to one (Fig. 72). We divide this square by a horizontal line into two equal parts and apply the upper part to the lower one so that a rectangle is formed with sides 2 and . After that, we again divide the right half of this rectangle in half by a horizontal line and attach the upper part to the lower one (as shown in Fig. 72). Continuing this process, we are constantly transforming the original square with area equal to 1 into equal-sized figures (taking the form of a staircase with thinning steps).

With an infinite continuation of this process, the entire area of ​​​​the square decomposes into an infinite number of terms - the areas of rectangles with bases equal to 1 and heights. The areas of the rectangles just form an infinite decreasing progression, its sum

i.e., as expected, is equal to the area of ​​the square.

Example. Find the sums of the following infinite progressions:

Solution, a) We note that this progression Therefore, by the formula (92.2) we find

b) Here it means that by the same formula (92.2) we have

c) We find that this progression Therefore, this progression has no sum.

In Section 5, the application of the formula for the sum of terms of an infinitely decreasing progression to the conversion of a periodic decimal fraction into an ordinary fraction was shown.

Exercises

1. The sum of an infinitely decreasing geometric progression is 3/5, and the sum of its first four terms is 13/27. Find the first term and denominator of the progression.

2. Find four numbers that form an alternating geometric progression, in which the second term is less than the first by 35, and the third is greater than the fourth by 560.

3. Show what if sequence

forms an infinitely decreasing geometric progression, then the sequence

for any form an infinitely decreasing geometric progression. Does this assertion hold for

Derive a formula for the product of the terms of a geometric progression.

Mathematics is whatpeople control nature and themselves.

Soviet mathematician, academician A.N. Kolmogorov

Geometric progression.

Along with tasks on arithmetic progressions, tasks related to the concept of a geometric progression are also common in entrance tests in mathematics. To successfully solve such problems, you need to know the properties of a geometric progression and have good skills in using them.

This article is devoted to the presentation of the main properties of a geometric progression. It also provides examples of solving typical problems, borrowed from the tasks of entrance tests in mathematics.

Let us preliminarily note the main properties of a geometric progression and recall the most important formulas and statements, associated with this concept.

Definition. A numerical sequence is called a geometric progression if each of its numbers, 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 a geometric progressionthe formulas are valid

, (1)

where . Formula (1) is called the formula of the general term of a geometric progression, and formula (2) is the main property of a geometric progression: each member of the progression coincides with the geometric mean of its neighboring members and .

Note, that it is precisely because of this property that the progression in question is called "geometric".

Formulas (1) and (2) above are summarized as follows:

, (3)

To calculate the sum first members of a geometric progressionthe formula applies

If we designate

where . Since , formula (6) is a generalization of formula (5).

In the case when and geometric progressionis infinitely decreasing. To calculate the sumof all members of an infinitely decreasing geometric progression, the formula is used

. (7)

For example , using formula (7), one can show, what

where . These equalities are obtained from formula (7) provided that , (the first equality) and , (the second equality).

Theorem. If , then

Proof. If , then ,

The theorem has been proven.

Let's move on to considering examples of solving problems on the topic "Geometric progression".

Example 1 Given: , and . To find .

Decision. If formula (5) is applied, then

Answer: .

Example 2 Let and . To find .

Decision. Since and , we use formulas (5), (6) and obtain the system of equations

If the second equation of system (9) is divided by the first, then or . From this it follows . Let's consider two cases.

1. If , then from the first equation of system (9) we have.

2. If , then .

Example 3 Let , and . To find .

Decision. It follows from formula (2) that or . Since , then or .

By condition . However , therefore . Because 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 single suitable root . In this case, the first equation of the system implies .

Taking into account formula (7), we obtain.

Answer: .

Example 4 Given: and . To find .

Decision. Since , then .

Because , 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 . To find .

Decision. According to the theorem, we have two equalities

Since , then or . Because , then .

Answer: .

Example 6 Given: and . To find .

Decision. Taking into account formula (5), we obtain

Since , then . Since , and , then .

Example 7 Let and . To find .

Decision. 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 .

Decision. From formula (7) it follows and . From here and from the condition of the problem, we obtain the 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.

Decision. 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 10solve the equation

, (11)

where and .

Decision. The left side of equation (11) is the sum of an infinite decreasing geometric progression, in which and , provided: and .

From formula (7) it follows, what . In this regard, equation (11) takes the form or . suitable root quadratic equation is

Answer: .

Example 11. P sequence of positive numbersforms an arithmetic progression, a - geometric progression, what does it have to do with . To find .

Decision. As arithmetic sequence, then (the main property of an arithmetic progression). Insofar as, then or . This implies , that the geometric progression is. According to formula (2), then we write that .

Since and , then . In that case, the expression takes the form or . By condition , so from the equationwe obtain the unique solution of the problem under consideration, i.e. .

Answer: .

Example 12. Calculate sum

. (12)

Decision. Multiply both sides of equality (12) by 5 and get

If we subtract (12) from the resulting expression, then

or .

To calculate, we substitute the values ​​into formula (7) and obtain . Since , then .

Answer: .

The examples of problem solving given here will be useful to applicants in preparation for entrance examinations. For a deeper study of problem solving methods, associated with a geometric progression, you can use the tutorials from the list of recommended literature.

1. Collection of tasks in mathematics for applicants to technical universities / Ed. M.I. Scanavi. – M.: Mir i Obrazovanie, 2013. – 608 p.

2. Suprun V.P. Mathematics for high school students: additional sections of the school curriculum. – M.: Lenand / URSS, 2014. - 216 p.

3. Medynsky M.M. A complete course of elementary mathematics in tasks and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. - 208 p.

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