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 invite you, in addition to the problems No. 995-1000 below, to 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.
A geometric progression is a numerical sequence, the first term of which is non-zero, and each next term is equal to the previous term multiplied by the same non-zero number.
The concept of geometric progression
The geometric progression is denoted by 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.
The sum of an infinite geometric progression for |q|<1
One way to set a geometric progression is to set its first term b1 and the denominator of the geometric error q. For example, b1=4, q=-2. These two conditions give a geometric progression of 4, -8, 16, -32, … .
If q>0 (q is not equal to 1), then the progression is a monotonic sequence. For example, the sequence, 2, 4,8,16,32, ... is a monotonically increasing sequence (b1=2, q=2).
If the denominator q=1 in the geometric error, then all members of the geometric progression will be equal to each other. In such cases, the progression is said to be a constant sequence.
In order for the numerical sequence (bn) to be a geometric progression, it is necessary that each of its members, starting from the second, be the geometric mean of the 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.
Now let's put (Xn) - a geometric progression. The denominator of the geometric progression q, with |q|∞).
If we now denote by S the sum of an infinite geometric progression, then the following formula will hold:
S=x1/(1-q).
Consider a simple example:
Find the sum of an infinite geometric progression 2, -2/3, 2/9, - 2/27, ... .
To find S, we use the formula for the sum of an infinitely arithmetic progression. |-1/3|< 1. x1 = 2. S=2/(1-(-1/3)) = 3/2.
If every natural number n match a real number a n , then they say that given number sequence :
a 1 , a 2 , a 3 , . . . , a n , . . . .
So, a numerical sequence is a function of a natural argument.
Number a 1 called the first member of the sequence , number a 2 — the second member of the sequence , number a 3 — third etc. Number a n called nth member of the sequence , and the natural number n — his number .
From two neighboring members a n and a n +1 member sequences a n +1 called subsequent (towards a n ), a a n — previous (towards a n +1 ).
To specify a sequence, you must specify a method that allows you to find a sequence member with any number.
Often the sequence is given with nth term formulas , that is, a formula that allows you to determine a sequence member by its number.
For example,
the sequence of positive odd numbers can be given by the formula
a n= 2n- 1,
and the sequence of alternating 1 and -1 - formula
b n = (-1)n +1 . ◄
The sequence can be determined recurrent formula, that is, a formula that expresses any member of the sequence, starting with some, through the previous (one or more) members.
For example,
if a 1 = 1 , a a n +1 = a n + 5
a 1 = 1,
a 2 = a 1 + 5 = 1 + 5 = 6,
a 3 = a 2 + 5 = 6 + 5 = 11,
a 4 = a 3 + 5 = 11 + 5 = 16,
a 5 = a 4 + 5 = 16 + 5 = 21.
If a a 1= 1, a 2 = 1, a n +2 = a n + a n +1 , then the first seven members of the numerical sequence are set as follows:
a 1 = 1,
a 2 = 1,
a 3 = a 1 + a 2 = 1 + 1 = 2,
a 4 = a 2 + a 3 = 1 + 2 = 3,
a 5 = a 3 + a 4 = 2 + 3 = 5,
a 6 = a 4 + a 5 = 3 + 5 = 8,
a 7 = a 5 + a 6 = 5 + 8 = 13. ◄
Sequences can be final and endless .
The sequence is called ultimate if it has a finite number of members. The sequence is called endless if it has infinitely many members.
For example,
sequence of two-digit natural numbers:
10, 11, 12, 13, . . . , 98, 99
final.
Prime number sequence:
2, 3, 5, 7, 11, 13, . . .
endless. ◄
The sequence is called increasing , if each of its members, starting from the second, is greater than the previous one.
The sequence is called waning , if each of its members, starting from the second, is less than the previous one.
For example,
2, 4, 6, 8, . . . , 2n, . . . is an ascending sequence;
1, 1 / 2 , 1 / 3 , 1 / 4 , . . . , 1 /n, . . . is a descending sequence. ◄
A sequence whose elements do not decrease with increasing number, or, conversely, do not increase, is called monotonous sequence .
Monotonic sequences, in particular, are increasing sequences and decreasing sequences.
Arithmetic progression
Arithmetic progression a sequence is called, each member of which, starting from the second, is equal to the previous one, to which the same number is added.
a 1 , a 2 , a 3 , . . . , a n, . . .
is an arithmetic progression if for any natural number n condition is met:
a n +1 = a n + d,
where d - some number.
Thus, the difference between the next and the previous members of a given arithmetic progression is always constant:
a 2 - a 1 = a 3 - a 2 = . . . = a n +1 - a n = d.
Number d called the difference of an arithmetic progression.
To set an arithmetic progression, it is enough to specify its first term and difference.
For example,
if a 1 = 3, d = 4 , then the first five terms of the sequence are found as follows:
a 1 =3,
a 2 = a 1 + d = 3 + 4 = 7,
a 3 = a 2 + d= 7 + 4 = 11,
a 4 = a 3 + d= 11 + 4 = 15,
a 5 = a 4 + d= 15 + 4 = 19. ◄
For an arithmetic progression with the first term a 1 and difference d her n
a n = a 1 + (n- 1)d.
For example,
find the thirtieth term of an arithmetic progression
1, 4, 7, 10, . . .
a 1 =1, d = 3,
a 30 = a 1 + (30 - 1)d= 1 + 29· 3 = 88. ◄
a n-1 = a 1 + (n- 2)d,
a n= a 1 + (n- 1)d,
a n +1 = a 1 + nd,
then obviously
| a n=
| a n-1 + a n+1
|
| 2
|
each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of the previous and subsequent members.
numbers a, b and c are consecutive members of some arithmetic progression if and only if one of them is equal to the arithmetic mean of the other two.
For example,
a n = 2n- 7 , is an arithmetic progression.
Let's use the statement above. We have:
a n = 2n- 7,
a n-1 = 2(n- 1) - 7 = 2n- 9,
a n+1 = 2(n+ 1) - 7 = 2n- 5.
Hence,
| a n+1 + a n-1
| =
| 2n- 5 + 2n- 9
| = 2n- 7 = a n,
|
| 2
| 2
|
◄
Note that n -th member of an arithmetic progression can be found not only through a 1 , but also any previous a k
a n = a k + (n- k)d.
For example,
for a 5 can be written
a 5 = a 1 + 4d,
a 5 = a 2 + 3d,
a 5 = a 3 + 2d,
a 5 = a 4 + d. ◄
a n = a n-k + kd,
a n = a n+k - kd,
then obviously
| a n=
| a n-k
+ a n+k
|
| 2
|
any member of an arithmetic progression, starting from the second, is equal to half the sum of the members of this arithmetic progression equally spaced from it.
In addition, for any arithmetic progression, the equality is true:
a m + a n = a k + a l,
m + n = k + l.
For example,
in arithmetic progression
1) a 10 = 28 = (25 + 31)/2 = (a 9 + a 11 )/2;
2) 28 = a 10 = a 3 + 7d= 7 + 7 3 = 7 + 21 = 28;
3) a 10= 28 = (19 + 37)/2 = (a 7 + a 13)/2;
4) a 2 + a 12 = a 5 + a 9, as
a 2 + a 12= 4 + 34 = 38,
a 5 + a 9 = 13 + 25 = 38. ◄
S n= a 1 + a 2 + a 3 + . . .+ a n,
first n members of an arithmetic progression is equal to the product of half the sum of the extreme terms by the number of terms:
From this, in particular, it follows that if it is necessary to sum the terms
a k, a k +1 , . . . , a n,
then the previous formula retains its structure:
For example,
in arithmetic progression 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, . . .
S 10 = 1 + 4 + . . . + 28 = (1 + 28) · 10/2 = 145;
10 + 13 + 16 + 19 + 22 + 25 + 28 = S 10 - S 3 = (10 + 28 ) · (10 - 4 + 1)/2 = 133. ◄
If an arithmetic progression is given, then the quantities a 1 , a n, d, n andS n linked by two formulas:
Therefore, if the values of three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas combined into a system of two equations with two unknowns.
An arithmetic progression is a monotonic sequence. Wherein:
- if d > 0 , then it is increasing;
- if d < 0 , then it is decreasing;
- if d = 0 , then the sequence will be stationary.
Geometric progression
geometric progression a sequence is called, each term of which, starting from the second, is equal to the previous one, multiplied by the same number.
b 1 , b 2 , b 3 , . . . , b n, . . .
is a geometric progression if for any natural number n condition is met:
b n +1 = b n · q,
where q ≠ 0 - some number.
Thus, the ratio of the next term of this geometric progression to the previous one is a constant number:
b 2 / b 1 = b 3 / b 2 = . . . = b n +1 / b n = q.
Number q called denominator of a geometric progression.
To set a geometric progression, it is enough to specify its first term and denominator.
For example,
if b 1 = 1, q = -3 , then the first five terms of the sequence are found as follows:
b 1 = 1,
b 2 = b 1 · q = 1 · (-3) = -3,
b 3 = b 2 · q= -3 · (-3) = 9,
b 4 = b 3 · q= 9 · (-3) = -27,
b 5 = b 4 · q= -27 · (-3) = 81. ◄
b 1 and denominator q her n -th term can be found by the formula:
b n = b 1 · q n -1 .
For example,
find the seventh term of a geometric progression 1, 2, 4, . . .
b 1 = 1, q = 2,
b 7 = b 1 · q 6 = 1 2 6 = 64. ◄
bn-1 = b 1 · q n -2 ,
b n = b 1 · q n -1 ,
b n +1 = b 1 · q n,
then obviously
b n 2 = b n -1 · b n +1 ,
each member of the geometric progression, starting from the second, is equal to the geometric mean (proportional) of the previous and subsequent members.
Since the converse is also true, the following assertion holds:
numbers a, b and c are consecutive members of some geometric progression if and only if the square of one of them is equal to the product of the other two, that is, one of the numbers is the geometric mean of the other two.
For example,
let us prove that the sequence given by the formula b n= -3 2 n , is a geometric progression. Let's use the statement above. We have:
b n= -3 2 n,
b n -1 = -3 2 n -1 ,
b n +1 = -3 2 n +1 .
Hence,
b n 2 = (-3 2 n) 2 = (-3 2 n -1 ) (-3 2 n +1 ) = b n -1 · b n +1 ,
which proves the required assertion. ◄
Note that n th term of a geometric progression can be found not only through b 1 , but also any previous term b k , for which it suffices to use the formula
b n = b k · q n - k.
For example,
for b 5 can be written
b 5 = b 1 · q 4 ,
b 5 = b 2 · q 3,
b 5 = b 3 · q2,
b 5 = b 4 · q. ◄
b n = b k · q n - k,
b n = b n - k · q k,
then obviously
b n 2 = b n - k· b n + k
the square of any member of a geometric progression, starting from the second, is equal to the product of the members of this progression equidistant from it.
In addition, for any geometric progression, the equality is true:
b m· b n= b k· b l,
m+ n= k+ l.
For example,
exponentially
1) b 6 2 = 32 2 = 1024 = 16 · 64 = b 5 · b 7 ;
2) 1024 = b 11 = b 6 · q 5 = 32 · 2 5 = 1024;
3) b 6 2 = 32 2 = 1024 = 8 · 128 = b 4 · b 8 ;
4) b 2 · b 7 = b 4 · b 5 , as
b 2 · b 7 = 2 · 64 = 128,
b 4 · b 5 = 8 · 16 = 128. ◄
S n= b 1 + b 2 + b 3 + . . . + b n
first n members of a geometric progression with a denominator q ≠ 0 calculated by the formula:
And when q = 1 - according to the formula
S n= n.b. 1
Note that if we need to sum the terms
b k, b k +1 , . . . , b n,
then the formula is used:
| S n- S k -1 = b k + b k +1 + . . . + b n = b k · | 1 - q n -
k +1
| . |
| 1 - q
|
For example,
exponentially 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, . . .
S 10 = 1 + 2 + . . . + 512 = 1 · (1 - 2 10) / (1 - 2) = 1023;
64 + 128 + 256 + 512 = S 10 - S 6 = 64 · (1 - 2 10-7+1) / (1 - 2) = 960. ◄
If a geometric progression is given, then the quantities b 1 , b n, q, n and S n linked by two formulas:
Therefore, if the values of any three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas combined into a system of two equations with two unknowns.
For a geometric progression with the first term b 1 and denominator q the following take place monotonicity properties :
- the progression is increasing if one of the following conditions is met:
b 1 > 0 and q> 1;
b 1 < 0 and 0 < q< 1;
- A progression is decreasing if one of the following conditions is met:
b 1 > 0 and 0 < q< 1;
b 1 < 0 and q> 1.
If a q< 0 , then the geometric progression is sign-alternating: its odd-numbered terms have the same sign as its first term, and even-numbered terms have the opposite sign. It is clear that an alternating geometric progression is not monotonic.
Product of the first n terms of a geometric progression can be calculated by the formula:
P n= b 1 · b 2 · b 3 · . . . · b n = (b 1 · b n) n / 2 .
For example,
1 · 2 · 4 · 8 · 16 · 32 · 64 · 128 = (1 · 128) 8/2 = 128 4 = 268 435 456;
3 · 6 · 12 · 24 · 48 = (3 · 48) 5/2 = (144 1/2) 5 = 12 5 = 248 832.◄
Infinitely decreasing geometric progression
Infinitely decreasing geometric progression is called an infinite geometric progression whose denominator modulus is less than 1 , i.e
|q| < 1 .
Note that an infinitely decreasing geometric progression may not be a decreasing sequence. This fits the case
1 < q< 0 .
With such a denominator, the sequence is sign-alternating. For example,
1, - 1 / 2 , 1 / 4 , - 1 / 8 , . . . .
The sum of an infinitely decreasing geometric progression name the number to which the sum of the first n terms of the progression with an unlimited increase in the number n . This number is always finite and is expressed by the formula
| S= b 1 + b 2 + b 3 + . . . = | b 1
| . |
| 1 - q
|
For example,
10 + 1 + 0,1 + 0,01 + . . . = 10 / (1 - 0,1) = 11 1 / 9 ,
10 - 1 + 0,1 - 0,01 + . . . = 10 / (1 + 0,1) = 9 1 / 11 . ◄
Relationship between arithmetic and geometric progressions
Arithmetic and geometric progressions are closely related. Let's consider just two examples.
a 1 , a 2 , a 3 , . . . d , then
b a 1 , b a 2 , b a 3 , . . . b d .
For example,
1, 3, 5, . . . — arithmetic progression with difference 2 and
7 1 , 7 3 , 7 5 , . . . is a geometric progression with a denominator 7 2 . ◄
b 1 , b 2 , b 3 , . . . is a geometric progression with a denominator q , then
log a b 1, log a b 2, log a b 3, . . . — arithmetic progression with difference log aq .
For example,
2, 12, 72, . . . is a geometric progression with a denominator 6 and
lg 2, lg 12, lg 72, . . . — arithmetic progression with difference lg 6 . ◄
Some problems of physics and mathematics can be solved using the properties of number series. The two simplest number sequences that are taught in schools are algebraic and geometric. In this article, we will consider in more detail the question of how to find the sum of an infinite progression of a geometric decreasing one.
geometric progression
These words mean such a series of real numbers, the elements a i of which satisfy the expression:
Here i is the number of the element in the series, r is a constant number, which is called the denominator.
This definition shows that, knowing any term of the progression and its denominator, it is possible to restore the entire series of numbers. For example, if the 10th element is known, then dividing it by r, we get the 9th element, then dividing it again, we get the 8th and so on. These simple arguments allow us to write an expression that is valid for the series of numbers under consideration:
An example of a progression with a denominator of 2 would be:
1, 2, 4, 8, 16, 32, ...
If the denominator is -2, then a completely different series is obtained:
1, -2, 4, -8, 16, -32, ...
A geometric progression is much faster than an algebraic one, that is, its terms increase rapidly and decrease rapidly.
The sum of i members of the progression
To solve practical problems, it is often necessary to calculate the sum of several elements of the considered numerical sequence. For this case, the following formula is valid:
S i \u003d a 1 * (r i -1) / (r-1)
It can be seen that to calculate the sum of i terms, you need to know only two numbers: a 1 and r, which is logical, since they uniquely determine the entire sequence.
Descending sequence and the sum of its terms
Now let's consider a special case. We will assume that the absolute value of the denominator r does not exceed one, i.e. -1 A decreasing geometric progression is interesting to consider because the infinite sum of its terms tends to a finite real number. Let's get the sum formula This is easy to do if we write out the expression for S i given in the previous paragraph. We have: S i \u003d a 1 * (r i -1) / (r-1) Consider the case when i->∞. Since the modulus of the denominator is less than 1, then raising it to an infinite power will give zero. This can be verified using the example r=0.5: 0,5 2 = 0,25; 0,5 3 = 0,125; ...., 0,5 20 = 0,0000009. As a result, the sum of the terms of an infinite geometric progression of decreasing will take the form: This formula is often used in practice, for example, to calculate the areas of figures. It is also used in solving the paradox of Zeno of Elea with a tortoise and Achilles. Obviously, considering the sum of an infinite progression of a geometric increasing (r>1), will lead to the result S ∞ = +∞. We will show how the above formulas should be applied using the example of solving the problem. It is known that the sum of an infinite geometric progression is 11. Moreover, its 7th term is 6 times less than the third term. What is the first element for this number series? First, let's write out two expressions for determining the 7th and 3rd elements. We get: Dividing the first expression by the second, and expressing the denominator, we have: a 7 / a 3 = r 4 => r = 4 √ (a 7 / a 3) Since the ratio of the seventh and third terms is given in the condition of the problem, we can substitute it and find r: r \u003d 4 √ (a 7 / a 3) \u003d 4 √ (1/6) ≈ 0.63894 We have calculated r with an accuracy of five significant digits after the decimal point. Since the resulting value is less than one, it means that the progression is decreasing, which justifies the use of the formula for its infinite sum. We write the expression for the first term in terms of the sum S ∞ : We substitute the known values \u200b\u200binto this formula and get the answer: a 1 \u003d 11 * (1-0.63894) \u003d 3.97166. Zeno of Elea is a famous Greek philosopher who lived in the 5th century BC. e. A number of its apogees or paradoxes have reached the present time, in which the problem of infinitely large and infinitely small in mathematics is formulated. One of Zeno's well-known paradoxes is the competition between Achilles and the tortoise. Zeno believed that if Achilles gave the tortoise some advantage in distance, he would never be able to overtake it. For example, let Achilles run 10 times faster than an animal crawling, which, for example, is 100 meters ahead of him. When the warrior runs 100 meters, the tortoise crawls back 10. Running 10 meters again, Achilles will see that the tortoise has crawled another 1 meter. You can argue like this indefinitely, the distance between the competitors will really decrease, but the turtle will always be in front. He led Zeno to the conclusion that movement does not exist, and all the surrounding movement of objects is an illusion. Of course, the ancient Greek philosopher was wrong. The solution to the paradox lies in the fact that an infinite sum of ever-decreasing segments tends to a finite number. In the above case, for the distance traveled by Achilles, we get: 100 + 10 + 1 + 0,1 + 0,01 + ... Applying the formula for the sum of an infinite geometric progression, we get: S ∞ \u003d 100 / (1-0.1) ≈ 111.111 meters This result shows that Achilles will overtake the tortoise when it crawls only 11.111 meters. The ancient Greeks did not know how to work with infinite quantities in mathematics. However, this paradox can be resolved if we pay attention not to the infinite number of gaps that Achilles must overcome, but to the finite number of steps the runner needs to achieve the goal. 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."
The problem of finding the first term of the progression
Zeno's famous paradox with the fast Achilles and the slow tortoise
