The general formula of geometric series is:
an=a 1 qn- 1
The first n terms and formulas are:
In geometric series, the proportional term:
And the relationship between any two terms am and an is an = amqn-m.
If the common ratio q of geometric progression satisfies 0 < ∣ q < ∣ 1, then this series is called infinite recursive geometric progression, which has various kinds.
The formula for the sum of items (also called the sum of all items) is:
From the definition of geometric series, the general term formula and the first n terms formula, we can deduce that:
a 1 an = a2 an- 1 = a3 an-2 =…= AK an-k+ 1，k∈{ 1，2，…，n}
If m, n, p, q∈N*, there are:
ap aq=am an，
Write π n = a 1 a2 ... Ann, and then there is.
π2n- 1=(an)2n- 1，π2n+ 1 =(an+ 1)2n+ 1
In addition, each term is a geometric series with positive numbers, and the same base number is taken to form a arithmetic progression; On the other hand, taking any positive number c as the cardinal number and a arithmetic progression term as the exponent, a power energy is constructed, which is a geometric series. In this sense, we say that a positive geometric series and an arithmetic series are isomorphic.
What matters is not only the definitions, properties and formulas of the two basic sequences; Moreover, the mathematical thinking and wisdom contained in the process of summation are extremely precious, such as "inverted addition" (arithmetic progression) and "dislocation subtraction" (geometric progression).
There are two main problems in series, one is to find the general term formula of series, and the other is to find the sum of the first n terms of series.
Third, examples.
Example 1. Let ap, aq, am and an be the first P, Q, M and N terms in the geometric series. If p+q=m+n, prove: apoaq=amoan.
It is proved that if the first term of geometric series is a 1 and the common ratio is q, then
ap=a 1 qp- 1，aq=a 1 qq- 1，am=a 1 qm- 1，an=a 1 qn- 1
So:
ap aq=a 12qp+q-2，am an=a 12 qm+n-2，
Therefore: AP AQ = am+an.
Note: This example is an important property of geometric series and is often used in solving problems. It shows that the product of two terms equidistant from both ends (the first two terms and the last two terms) in geometric series is equal to the product of the first two terms and the last two terms, namely:
a 1+k an-k=a 1 an
The same is true for arithmetic progression: in arithmetic progression, the sum of two terms, such as the distance from both ends, is equal to the sum of the first term and the last term. Namely:
a 1+k+an-k=a 1+an
Example 2. In arithmetic progression, a4+a6+A8+a10+a12 =120, then 2a9-a 10=
a20 b . 22 c . 24 D28
Solution: a4+a 12=2a8, a6+a 10 =2a8 and known or obtained.
5a8= 120，a8=24
And 2a9-a10 = 2 (a1+8d)-(a1+9d) = a1+7d = A8 = 24.
So choose C.
Example 3. Given that arithmetic progression satisfies a1+a2+a3+…+a10/= 0, there is ().
a . a 1+a 10 1 > 0 b . a2+a 100 < 0 c . a3+a99 = 0d . a 5 1 = 5 1
[2000 Beijing Spring College Entrance Examination of Science and Engineering (13)]
Solution: Obviously, a1+A2+A3+…+A10/.
So a 1+a 10 1=0, so a2+a100 = a3+a99 = a1+a1= 0.
Example 4. Let Sn be the first n term of arithmetic progression, S9= 18, An-4 = 30 (n > 9), and Sn=336, then n is ().
A. 16 B.2 1 C.9 D8
Solution: Because S9=9×a5= 18, a5=2, a5+an-4=a 1+an=2+30=32, therefore, n=2 1 choose B.
Example 5. Let arithmetic progression satisfy 3a8=5a 13, and a 1>0 > 0, and Sn is the sum of its first n terms, then the largest of Sn(n∈N*) is (). (1995 National Senior High School LeagueNo. 1)
(A)s 10(B)s 1 1(C)s 20(D)s 2 1
Solution: ∫3 A8 = 5a 13
∴3(a 1+7d)=5(a 1+ 12d)
therefore
Make an ≥ 0 → n ≤ 20; When n > 20, An < 0.
∴S 19=S20 maximum, select (c)
Note: Quadratic function can also be used to find the maximum value.
Example 6. Let the first sum tolerance of arithmetic progression be a non-negative integer, the number of terms is not less than 3, and the sum of terms is 972, then such a series * * * has ().
2 (B)3 (C)4 (D)5。
[1997 The Third Question of the National Senior High School Mathematics League]
Solution: Let the first term of arithmetic progression be A and the tolerance be D, then there is () according to the meaning of the question.
That is, [2a+(n- 1)d]on=2×972 (*)
Because n is a natural number not less than 3 and 97 is a prime number, the value of the number n must be a divisor (factor) of 2×972, which can only be one of 97, 2×97, 972 and 2×972.
If d > 0, then d≥ 1 can be known from formula (*) as 2× 972 ≥ n (n-1) and d ≥ n (n-1), so there can only be n=97, and formula (*) can be changed to: a+48d.
If d=0, the formula (*) becomes: an=972, then (*) also has two sets of solutions.
So there are four arithmetic progression * * * set the conditions for this problem, namely:
49, 50, 5 1, …, 145, (***97 items)
1, 3, 5, …, 193, (**97 items)
97, 97, 97, …, 97, (**97 items)
1, 1, 1, …, 1(***972=9409 items)
So choose (c)
Example 7. Will be odd set {1, 3, 5, ...} according to the (2n- 1) odd group from small to large:
, {3,5,7},{9, 1 1, 13, 15, 17},…
(Group 1) (Group 2) (Group 3)
Then 199 1 is in the group.
[199 1 the third question of the national high school mathematics league]
Solution: according to the meaning of the question, the first n groups have odd numbers.
1+3+5+…+(2n- 1)=n2。
And1991= 2× 996-1is the 996th positive odd number.
∵3 12=96 1 0, 0 < x-[x] < 1, then the solution is:
From 0 < x-[x] < 1,
∴[x]= 1，
Therefore, it should be filled in
Example 9. The first term of geometric series a 1= 1536, common ratio. If πn is used to represent the product of the first n terms, then the largest πn(n∈N*) is ().
(A)π9(B)π 1 1(C)π 12(D)π 13
[1996 National Senior High School Mathematics League Examination Questions]
Solution: The general formula of geometric series is the sum of the first n terms.
because
So π 12 is the largest.
Option (c)
Example 10. Let x≠y and two series x, a 1, a2, a3, y and b 1, x, b2, b3, y and b4 be arithmetic progression, then =
[1988 National Senior High School League Examination Questions]
Solution: According to the meaning of the question, y-x = 4 (A2-a1) ∴;
And y-x = 3 (B3-B2) VII.
∴
Example 1 1. Let x, y, Z y and z be real numbers, and 3x, 4y and 5z be geometric progression and arithmetic progression, then the value of is. [1992 National Senior High School Mathematics League Examination Questions]
Solution: Because 3x, 4y and 5z are geometric series, there is.
3x 5z = (4y) 2, that is, 16y2= 15xz ①.
And ∵ into arithmetic progression, so there is (2).
Substitute ② into ① to get:
∵x≠0，y≠0，z≠0
∴64xz= 15(x2+2xz+z2)
∴ 15(x2+z2)=34xz
∴
Example 12. The known sets M={x, xy, lg(xy)} and N={0,∣x∣,y}
And M=N, the value of is equal to.
Solution: From M=N, we know that an element in m should be 0, so that lg(xy) can know xy≠0 meaningfully, thus x≠0 and y≠0, so only lg(xy)=0, xy= 1, m = {x,/kl. If y= 1, then x= 1, m = n = {0, 1, 1} is connected with the anisotropy of elements in the set, so y≠ 1, thus ∣ x ∣. . X= 1 y= 1 (inclusive), x=- 1 y=- 1, M=N={0, 1,-1}
At this moment,
therefore
Note: x, x2, x3, …, X200 1 series; and
Under the condition of x=y=- 1, they are all cyclic sequences with a period of 2, S2n- 1=-2 and S2n=0, so 200 1 is not terrible.
Example 13. If the known sequence satisfies 3an+ 1+an=4(n≥ 1) and a 1=9, and the sum of the first n terms is Sn, then inequality () is satisfied.
∣ sn-n-6 ∣ 0, ∴. Therefore, for any 1≤k≤n, there is
Note S = a11+A22+A33+…+ANN ⑤
⑥
⑤-⑤:
that is
Comments: The sum in this question is actually the sum of the first n items of the new series formed by the product of arithmetic progression's an=n and geometric progression's corresponding items. Formula ⑤ Multiplies the common ratio on both sides, and then subtracts the wrong term, which comes down to the solution of geometric series. This method is the basic method to find the sum of the first n terms of geometric series, which is very useful in solving this kind of problems and should be mastered. Textbook P 137 Review Reference Question 3 B Group 6 is entitled: SUM: S =1+2x+3x2+…+NXN-1; Problems of Science and Engineering in Beijing College Entrance Examination in 2003 (16): The known series is arithmetic progression, A 1 = 2, A 1+A2+A3 = 12; (1) Find the general formula of the series; (II) Let BN = an xn (x ∈ r), and find the summation formula of the first n items in the sequence. All of them run through the application of the method of "subtraction of wrong questions"
practise
1. Given the geometric series with q (q≠ 1), let b 1=a 1+a2+a3, b2=a4+a5+a6, …, BN = A3N-2+A3N-655.
Arithmetic series; (b) the geometric series with the common ratio q.
(c) Geometric series with a common ratio of q3; (d) arithmetic progression and geometric series.
[1999 National High School Mathematics Competition]
2. The sum of the first m terms of arithmetic progression is 30, and the sum of the first 2m terms is 100, so the sum of its first 3m terms is ().
130 b . 170 c . 2 10d . 260
[1996 National College Entrance Examination]
3. In arithmetic progression, a 1=2, the tolerance is not zero, and a 1, a3, a1/are just the first three terms of a geometric progression, so the value of the common ratio of this geometric progression is equal to.
[2002 Beijing College Entrance Examination for Science and Technology Mathematics 14]
4. The known sequence is arithmetic progression, and a 1=2, a 1+a2+a3= 12.
(i) Find the general term formula of the sequence;
(II) (text) let BN = an 3n, and find the formula of the sum of the first n items in the sequence;
Let bn = an xn (x ∈ r), and find a formula for the sum of the first n terms of a series.
[Question 16 of the Mathematics of the 2003 Beijing Summer College Entrance Examination]
5. Total:
( 1)S = 1+2x+3 x2+…+nxn- 1
[Mathematics Textbook Volume 1 (1) P 137 Review Reference Question 3 B Group Question 6]
(2) Find the sum of the first n items in the sequence Sn: 1, 6,27, …, n-3n- 1.
6. It is known that the positive integer n does not exceed 2000 and can be expressed as the sum of not less than 60 consecutive positive integers, so the number of such n is [1999 National Senior High School Mathematics Competition].
7. The tolerance of arithmetic progression whose term is a real number is 4, and the sum of squares of the first term and other terms does not exceed 100. This series has at most terms. [1998 National High School Mathematics Competition]
Reference answer
1. (3)
2. (3)
3.4
4.(I)an=2n
㈡
5.
6.6
7.8