Geometric progression is one of the most interesting number series considered in the school algebra course. This article is devoted to a special case of the mentioned series: a decreasing infinite geometric progression and the sum of its members.
What series of numbers will we talk about?
Geometric progression is a one-dimensional sequence of real numbers that are related to each other by the following relation:
a 2 = a 1 * r, a 3 = a 2 * r, a 4 = a 3 * r, ...., a n = a n-1 * r
Summarizing the expressions above, we can write the following equality:
a n = a 1 * r n-1
As is clear from the above records, a n is an element of the progression with the number n. The parameter r, by which the n-1 element should be multiplied to get the nth, is called the denominator.
What properties does the described sequence have? The answer to the question depends on the magnitude and sign of r. The following options are possible:
- The denominator r is positive and greater than 1. Progression in this case will always increase in absolute value, while the absolute value of its members may decrease if a 1 is negative.
- The denominator r is negative and greater than 1. In this case, the members of the progression will appear with alternating signs (+ and -). Such series are of little interest for practice.
- The denominator modulus r is less than 1. This series is called decreasing, and regardless of the sign of r. It is this progression that is of great practical interest, which will be discussed in this article.
Formula for amount
First we get an expression that allows us to calculate the sum of an arbitrary number of elements of a given progression. Let's start solving this problem head on. We have:
S n = a 1 + a 2 + a 3 + .. + a n
The above equality can be used if it is necessary to calculate the result for a small number of terms (3-4 terms), each of which is determined by the formula for the nth term (see the previous paragraph). However, if there are a lot of terms, then it is inconvenient to consider it in the forehead and a mistake can be made, therefore, they use a special formula.
We multiply both sides of the equality by r, we get:
r * S n = r * a 1 + r * a 2 + r * a 3 + .. + r * a n = a 2 + a 3 + a 4 + ... + a n + 1
Now subtract the left and right sides of these two expressions in pairs, we have:
r * S n - S n = a 2 + a 3 + a 4 + ... + a n + 1 - (a 1 + a 2 + a 3 + .. + a n ) = a n + 1 - a 1
Expressing the sum S n and using the formula for the term a n + 1 , we obtain:
S n = (a n + 1 - a 1 ) / (r-1) = a 1 * (r n - 1) / (r-1)
Thus, we have obtained a general formula for the sum of the first n terms of the considered type of number series. Note that the formula is valid if r ≠ 1. In the latter case, there is a simple series of identical numbers, the sum of which is calculated as the product of one number and their number.
How to find the sum of an infinite geometric progression decreasing?
To answer this question, it should be recalled that the series will be decreasing when | r | <1. We use the formula obtained for the previous paragraph for S n :
S n = a 1 * (r n - 1) / (r-1)
Note that any number whose modulus is less than 1 tends to zero when raised to a large degree, that is, r ∞ -> 0. You can verify this fact with any example:
r = -1/2, then (-1/2) ** 10 ≈ 9.7 * 10 -4 , (-1/2) ** 20 ≈ 9.5 * 10 -7 and so on.
Having established this fact, we pay attention to the expression for the sum: for n-> ∞, it will be rewritten as follows:
S ∞ = a 1 * (r ∞ - 1) / (r-1) = a 1 / (1-r)
An interesting result was obtained: the sum of an infinite progression of a geometric decreasing tends to a finite number, which does not depend on the number of terms. It is determined only by the first term and denominator. Note that the sign of the sum is uniquely determined by the sign of a 1 , since the denominator is always a positive number (1-r> 0).
Sum of squares of infinite geometric progression decreasing
The name of the item determines the task to be solved. To do this, we use a technique that is completely similar to that used to derive the general formula for S n . We have the first expression:
M n = a 1 2 + a 2 2 + a 3 2 + ... + a n 2
We multiply both sides of the equality by r 2 , write the second expression:
r 2 * M n = r 2 * a 1 2 + r 2 * a 2 2 + r 2 * a 3 2 + ... + r 2 * a n 2 = a 2 2 + a 3 2 + a 4 2 . .. + a n + 1 2
Now we find the difference of these two equalities:
r 2 * M n - M n = a 2 2 + a 3 2 + a 4 2 ... + a n + 1 2 - (a 1 2 + a 2 2 + a 3 2 + ... + a n 2 ) = a n + 1 2 - a 1 2
We express M n and use the formula for the nth element, we obtain the equality:
M n = (a n + 1 2 - a 1 2 ) / (r 2 -1) = a 1 2 * (r 2n -1) / (r 2 -1)
In the previous paragraph, it was shown that r ∞ -> 0, then the final formula takes the form:
M ∞ = a 1 2 * / (1-r 2 )
Comparison of two received amounts
Let us compare two formulas: for an infinite sum and an infinite sum of squares using the example of the following problem: the sum of an infinite geometric progression is 2, it is known that we are talking about a decreasing sequence for which the denominator is 1/3. It is necessary to find the infinite sum of squares of this series of numbers.
We use the formula for the sum. Express a 1 :
S ∞ = a 1 / (1-r) => a 1 = S ∞ * (1-r)
We substitute this expression into the formula for the sum of squares, we have:
M ∞ = a 1 2 * / (1-r 2 ) = S ∞ 2 * (1-r) 2 / (1-r 2 ) = S ∞ 2 * (1-r) / (1 + r)
We got the desired formula, now we can substitute the data known from the condition:
M ∞ = S ∞ 2 * (1-r) / (1 + r) = 2 2 * (1-1 / 3) / (1 + 1/3) = 2
Thus, we have obtained the same value for an infinite sum of squares as for a simple sum. Note that this result is valid only for this problem. In the general case, M ∞ ≠ S ∞ .
The task of calculating the area of a rectangle
Each student knows the formula S = a * b, which determines the area of the rectangle through its sides. Few people know that the problem of finding the area of this figure can be easily solved by using the sum of an infinite geometric progression. We show how this is done.
Mentally divide the rectangle in half. We take the area of one half per unit. Now divide the other half in half. We get two halves, one of which we will divide in half. We will continue this procedure ad infinitum (see the figure below).
As a result, the area of the rectangle in units selected by us will be equal to:
S ∞ = 1 + 1/2 + 1/4 + 1/8 + ...
It can be seen that these terms are elements of a decreasing series for which a 1 = 1 and r = 1/2. Using the formula for an infinite sum, we get:
S ∞ = 1 / (1-1 / 2) = 2
At our chosen scale, half of the rectangle (one unit) corresponds to the area a * b / 2. This means that the area of the entire rectangle is:
S ∞ = 2 * a * b / 2 = a * b
The result obtained is obvious, however, he showed how decreasing progression can be applied to solve problems in geometry.