The formula for the sum of geometric progressions. Geometric progression

The purpose of the lesson: to introduce students to a new kind of sequence - an infinitely decreasing geometric progression.
Tasks:
formulation of the initial idea of ​​the limit of the numerical sequence;
acquaintance with another way of converting infinite periodic fractions into ordinary ones using the formula for the sum of an infinitely decreasing geometric progression;
the development of the intellectual qualities of the personality of schoolchildren, such as logical thinking, the ability for evaluative actions, generalization;
education of activity, mutual assistance, collectivism, interest in the subject.

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Related lesson “Infinitely decreasing geometric progression” (algebra, grade 10)

The purpose of the lesson: introducing students to a new kind of sequence - an infinitely decreasing geometric progression.

Tasks:

formulation of the initial idea of ​​the limit of the numerical sequence; acquaintance with another way of converting infinite periodic fractions into ordinary ones using the formula for the sum of an infinitely decreasing geometric progression;

the development of the intellectual qualities of the personality of schoolchildren, such as logical thinking, the ability for evaluative actions, generalization;

education of activity, mutual assistance, collectivism, interest in the subject.

Equipment: computer class, projector, screen.

Lesson type: Lesson - mastering a new topic.

During the classes

I. Org. moment. Message about the topic and purpose of the lesson.

II. Updating students' knowledge.

In 9th grade, you studied arithmetic and geometric progressions.

Questions

1. Definition of an arithmetic progression.

(An arithmetic progression is a sequence in which each member,

Starting from the second, it is equal to the previous term, added with the same number).

2. Formula n -th member of an arithmetic progression

3. The formula for the sum of the first n members of an arithmetic progression.

( or )

4. Definition of a geometric progression.

(A geometric progression is a sequence of non-zero numbers,

Each term of which, starting from the second, is equal to the previous term, multiplied by

the same number).

5. Formula n th term of a geometric progression

6. The formula for the sum of the first n members of a geometric progression.

7. What formulas do you still know?

(, where ; ;

; , )

Tasks

1. Arithmetic progression is given by the formula a n = 7 - 4n. Find a 10 . (-33)

2. Arithmetic progression a 3 = 7 and a 5 = 1 . Find a 4 . (4)

3. Arithmetic progression a 3 = 7 and a 5 = 1 . Find a 17 . (-35)

4. Arithmetic progression a 3 = 7 and a 5 = 1 . Find S 17 . (-187)

5. For a geometric progressionfind the fifth term.

6. For a geometric progression find the nth term.

7. Exponentially b 3 = 8 and b 5 = 2 . Find b 4 . (4)

8. Exponentially b 3 = 8 and b 5 = 2 . Find b 1 and q .

9. Exponentially b 3 = 8 and b 5 = 2 . Find S 5 . (62)

III. Exploring a new topic(demonstration presentation).

Consider a square with a side equal to 1. Let's draw another square, the side of which is half the first square, then another one, the side of which is half the second, then the next one, and so on. Each time the side of the new square is half the previous one.

As a result, we got a sequence of sides of squaresforming a geometric progression with a denominator.

And, what is very important, the more we build such squares, the smaller the side of the square will be. For example ,

Those. as the number n increases, the terms of the progression approach zero.

With the help of this figure, one more sequence can be considered.

For example, the sequence of areas of squares:

And, again, if n increases indefinitely, then the area approaches zero arbitrarily close.

Let's consider one more example. An equilateral triangle with a side of 1 cm. Let's construct the next triangle with vertices in the midpoints of the sides of the 1st triangle, according to the triangle midline theorem - the side of the 2nd is equal to half the side of the first, the side of the 3rd is half the side of the 2nd, etc. Again we get a sequence of lengths of the sides of the triangles.

At .

If we consider a geometric progression with a negative denominator.

Then, again, with increasing numbers n the terms of the progression approach zero.

Let's pay attention to the denominators of these sequences. Everywhere the denominators were less than 1 modulo.

We can conclude: a geometric progression will be infinitely decreasing if the modulus of its denominator is less than 1.

Front work.

Definition:

A geometric progression is said to be infinitely decreasing if the modulus of its denominator is less than one..

With the help of the definition, it is possible to solve the question of whether a geometric progression is infinitely decreasing or not.

Task

Is the sequence an infinitely decreasing geometric progression if it is given by the formula:

Decision:

Let's find q .

; ; ; .

this geometric progression is infinitely decreasing.

b) this sequence is not an infinitely decreasing geometric progression.

Consider a square with a side equal to 1. Divide it in half, one of the halves in half again, and so on. the areas of all the resulting rectangles form an infinitely decreasing geometric progression:

The sum of the areas of all the rectangles obtained in this way will be equal to the area of ​​the 1st square and equal to 1.

But on the left side of this equality is the sum of an infinite number of terms.

Consider the sum of the first n terms.

According to the formula for the sum of the first n terms of a geometric progression, it is equal to.

If n increases indefinitely, then

or . Therefore, i.e. .

The sum of an infinitely decreasing geometric progressionthere is a sequence limit S 1 , S 2 , S 3 , …, S n , … .

For example, for a progression,

we have

As

The sum of an infinitely decreasing geometric progressioncan be found using the formula.

III. Reflection and Consolidation(completion of tasks).

№13; №14; №15(1,3); №16(1,3); №18(1,3); №19; №20.

IV. Summarizing.

What sequence did you meet today?

Define an infinitely decreasing geometric progression.

How to prove that a geometric progression is infinitely decreasing?

Give the formula for the sum of an infinitely decreasing geometric progression.

V. Homework.

2. № 15(2,4); №16(2,4); 18(2,4).

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Slides captions:

Everyone should be able to think consistently, judge conclusively, and refute wrong conclusions: a physicist and a poet, a tractor driver and a chemist. E.Kolman In mathematics, one should remember not formulas, but processes of thinking. VP Ermakov It is easier to find the square of a circle than to outwit a mathematician. Augustus de Morgan What science could be more noble, more admirable, more useful to mankind than mathematics? Franklin

Infinitely decreasing geometric progression Grade 10

I. Arithmetic and geometric progressions. Questions 1. Definition of an arithmetic progression. An arithmetic progression is a sequence in which each term, starting from the second, is equal to the previous term added to the same number. 2. Formula of the nth member of an arithmetic progression. 3. The formula for the sum of the first n members of an arithmetic progression. 4. Definition of a geometric progression. A geometric progression is a sequence of non-zero numbers, each member of which, starting from the second, is equal to the previous member multiplied by the same number 5. The formula of the nth member of a geometric progression. 6. The formula for the sum of the first n members of a geometric progression.

II. Arithmetic progression. Assignments An arithmetic progression is given by the formula a n = 7 – 4 n Find a 10 . (-33) 2. In arithmetic progression a 3 = 7 and a 5 = 1 . Find a 4 . (4) 3. In arithmetic progression a 3 = 7 and a 5 = 1 . Find a 17 . (-35) 4. In arithmetic progression a 3 = 7 and a 5 = 1 . Find S 17 . (-187)

II. Geometric progression. Tasks 5. For a geometric progression, find the fifth term 6. For a geometric progression, find the n-th term. 7. Exponentially b 3 = 8 and b 5 = 2. Find b 4 . (4) 8. In geometric progression b 3 = 8 and b 5 = 2 . Find b 1 and q . 9. In geometric progression b 3 = 8 and b 5 = 2. Find S 5 . (62)

definition: A geometric progression is said to be infinitely decreasing if the modulus of its denominator is less than one.

Problem №1 Is the sequence an infinitely decreasing geometric progression, if it is given by the formula: Solution: a) this geometric progression is infinitely decreasing. b) this sequence is not an infinitely decreasing geometric progression.

The sum of an infinitely decreasing geometric progression is the limit of the sequence S 1 , S 2 , S 3 , …, S n , … . For example, for a progression, we have Since the sum of an infinitely decreasing geometric progression can be found by the formula

Completion of tasks Find the sum of an infinitely decreasing geometric progression with the first term 3, the second 0.3. 2. No. 13; No. 14; textbook, p. 138 3. No. 15 (1; 3); #16(1;3) #18(1;3); 4. No. 19; No. 20.

What sequence did you meet today? Define an infinitely decreasing geometric progression. How to prove that a geometric progression is infinitely decreasing? Give the formula for the sum of an infinitely decreasing geometric progression. Questions

The famous Polish mathematician Hugo Steinghaus jokingly claims that there is a law that is formulated as follows: a mathematician will do it better. Namely, if you entrust two people, one of whom is a mathematician, to do any work they do not know, then the result will always be the following: the mathematician will do it better. Hugo Steinghaus 14.01.1887-25.02.1972

This number is called the denominator of a geometric progression, that is, 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 a 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."

Geometric progression no less important in mathematics than in arithmetic. A geometric progression is such a sequence of numbers b1, b2,..., b[n] each next member of which is obtained by multiplying the previous one by a constant number. This number, which also characterizes the rate of growth or decrease of the progression, is called denominator of a geometric progression and denote

For a complete assignment of a geometric progression, in addition to the denominator, it is necessary to know or determine its first term. For a positive value of the denominator, the progression is a monotone sequence, and if this sequence of numbers is monotonically decreasing and monotonically increasing when. The case when the denominator is equal to one is not considered in practice, since we have a sequence of identical numbers, and their summation is not of practical interest

General term of a geometric progression calculated according to the formula

The sum of the first n terms of a geometric progression determined by the formula

Let us consider solutions of classical geometric progression problems. Let's start with the simplest to understand.

Example 1. The first term of a geometric progression is 27, and its denominator is 1/3. Find the first six terms of a geometric progression.

Solution: We write the condition of the problem in the form

For calculations, we use the formula for the nth member of a geometric progression

Based on it, we find unknown members of the progression

As you can see, calculating the terms of a geometric progression is not difficult. The progression itself will look like this

Example 2. The first three members of a geometric progression are given: 6; -12; 24. Find the denominator and the seventh term.

Solution: We calculate the denominator of the geometric progression based on its definition

We got an alternating geometric progression whose denominator is -2. The seventh term is calculated by the formula

On this task is solved.

Example 3. A geometric progression is given by two of its members . Find the tenth term of the progression.

Decision:

Let's write the given values ​​​​through the formulas

According to the rules, it would be necessary to find the denominator, and then look for the desired value, but for the tenth term we have

The same formula can be obtained on the basis of simple manipulations with the input data. We divide the sixth term of the series by another, as a result we get

If the resulting value is multiplied by the sixth term, we get the tenth

Thus, for such problems, with the help of simple transformations in a fast way, you can find the right solution.

Example 4. Geometric progression is given by recurrent formulas

Find the denominator of the geometric progression and the sum of the first six terms.

Decision:

We write the given data in the form of a system of equations

Express the denominator by dividing the second equation by the first

Find the first term of the progression from the first equation

Compute the following five terms to find the sum of the geometric progression

Let's consider a series.

7 28 112 448 1792...

It is absolutely clear that the value of any of its elements is exactly four times greater than the previous one. So this series is a progression.

A geometric progression is an infinite sequence of numbers, the main feature of which is that the next number is obtained from the previous one by multiplying by some specific number. This is expressed by the following formula.

a z +1 =a z q, where z is the number of the selected element.

Accordingly, z ∈ N.

The period when a geometric progression is studied at school is grade 9. Examples will help you understand the concept:

0.25 0.125 0.0625...

Based on this formula, the denominator of the progression can be found as follows:

Neither q nor b z can be zero. Also, each of the elements of the progression should not be equal to zero.

Accordingly, to find out the next number in the series, you need to multiply the last one by q.

To specify this progression, you must specify its first element and denominator. After that, it is possible to find any of the subsequent terms and their sum.

Varieties

Depending on q and a 1, this progression is divided into several types:

• If both a 1 and q are greater than one, then such a sequence is a geometric progression increasing with each next element. An example of such is presented below.

Example: a 1 =3, q=2 - both parameters are greater than one.

Then the numerical sequence can be written like this:

3 6 12 24 48 ...

• If |q| less than one, that is, multiplication by it is equivalent to division, then a progression with similar conditions is a decreasing geometric progression. An example of such is presented below.

Example: a 1 =6, q=1/3 - a 1 is greater than one, q is less.

Then the numerical sequence can be written as follows:

6 2 2/3 ... - any element is 3 times greater than the element following it.

• Sign-variable. If q<0, то знаки у чисел последовательности постоянно чередуются вне зависимости от a 1 , а элементы ни возрастают, ни убывают.

Example: a 1 = -3 , q = -2 - both parameters are less than zero.

Then the sequence can be written like this:

3, 6, -12, 24,...

Formulas

For convenient use of geometric progressions, there are many formulas:

• Formula of the z-th member. Allows you to calculate the element under a specific number without calculating the previous numbers.

Example:q = 3, a 1 = 4. It is required to calculate the fourth element of the progression.

Decision:a 4 = 4 · 3 4-1 = 4 · 3 3 = 4 · 27 = 108.

• The sum of the first elements whose number is z. Allows you to calculate the sum of all elements of a sequence up toa zinclusive.

Since (1-q) is in the denominator, then (1 - q)≠ 0, hence q is not equal to 1.

Note: if q=1, then the progression would be a series of an infinitely repeating number.

The sum of a geometric progression, examples:a 1 = 2, q= -2. Calculate S 5 .

Decision:S 5 = 22 - calculation by formula.

• Amount if |q| < 1 и если z стремится к бесконечности.

Example:a 1 = 2 , q= 0.5. Find the amount.

Decision:Sz = 2 · = 4

Sz = 2 + 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 3.9375 4

Some properties:

• characteristic property. If the following condition performed for anyz, then the given number series is a geometric progression:

a z 2 = a z -1 · az+1

• Also, the square of any number of a geometric progression is found by adding the squares of any other two numbers in a given series, if they are equidistant from this element.

a z 2 = a z - t 2 + a z + t 2 , wheretis the distance between these numbers.

• Elementsdiffer in qonce.
• The logarithms of the progression elements also form a progression, but already arithmetic, that is, each of them is greater than the previous one by a certain number.

Examples of some classical problems

To better understand what a geometric progression is, examples with a solution for grade 9 can help.

• Conditions:a 1 = 3, a 3 = 48. Findq.

Solution: each subsequent element is greater than the previous one inq once.It is necessary to express some elements through others using a denominator.

Hence,a 3 = q 2 · a 1

When substitutingq= 4

• Conditions:a 2 = 6, a 3 = 12. Calculate S 6 .

Decision:To do this, it is enough to find q, the first element and substitute it into the formula.

a 3 = q· a 2 , hence,q= 2

a 2 = q a 1 ,That's why a 1 = 3

S 6 = 189

• · a 1 = 10, q= -2. Find the fourth element of the progression.

Solution: to do this, it is enough to express the fourth element through the first and through the denominator.

a 4 = q 3· a 1 = -80

Application example:

• The client of the bank made a deposit in the amount of 10,000 rubles, under the terms of which every year the client will add 6% of it to the principal amount. How much money will be in the account after 4 years?

Solution: The initial amount is 10 thousand rubles. So, a year after the investment, the account will have an amount equal to 10,000 + 10,000 · 0.06 = 10000 1.06

Accordingly, the amount in the account after another year will be expressed as follows:

(10000 1.06) 0.06 + 10000 1.06 = 1.06 1.06 10000

That is, every year the amount increases by 1.06 times. This means that in order to find the amount of funds in the account after 4 years, it is enough to find the fourth element of the progression, which is given by the first element equal to 10 thousand, and the denominator equal to 1.06.

S = 1.06 1.06 1.06 1.06 10000 = 12625

Examples of tasks for calculating the sum:

In various problems, a geometric progression is used. An example for finding the sum can be given as follows:

a 1 = 4, q= 2, calculateS5.

Solution: all the data necessary for the calculation are known, you just need to substitute them into the formula.

S 5 = 124

• a 2 = 6, a 3 = 18. Calculate the sum of the first six elements.

Decision:

Geom. progression, each next element is q times greater than the previous one, that is, to calculate the sum, you need to know the elementa 1 and denominatorq.

a 2 · q = a 3

q = 3

Similarly, we need to finda 1 , knowinga 2 andq.

a 1 · q = a 2

a 1 =2

S 6 = 728.

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 an 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.