1. in arithmetic progression {an}, if a1+a2+a12+a13 = 24, then a7 is ().
a6 b . 7 c . 8d . 9
Analysis: ∫ a1+a2+a12+a13 = 4a7 = 24, ∴ A7 = 6.
A: A.
2. If the sum of the first n terms of arithmetic progression {an} is Sn, and S33-S22 = 1, the tolerance of sequence {an} is ().
A. 12 B. 1 C.2 D.3
Analysis: from sn = na 1+n (n- 1) 2d, we get S3 = 3a 1+3d, S2 = 2a 1+d, and substitute s33-s22 = 1 to get d = 2, so we choose.
Answer: c
3. Given the sequence A 1 = 1, A2 = 5, An+2 = An+ 1-An (n ∈ n *), a2 0 1 1 equals ().
a . 1 B- 4 c . 4d . 5
Analysis: From the known, A 1 = 1, A2 = 5, A3 = 4, A4 =- 1, A5 =-5, A6 =-4, A7 = 1, A8 = 5, …
So {an} is a series with a period of 6.
∴a2 0 1 1 = a6×335+ 1 = a 1 = 1。
A: A.
4. Let {an} be arithmetic progression, Sn be the sum of its first n terms, S5 < S6, S6 = S7 > S8, then the following conclusion is wrong ().
A.d S5 D. S6 and S7 are the maximum values of Sn.
Analysis: ∵ S5 < S6, ∴ A6 > 0. S6 = S7，∴ A7 = 0。
S7 > S8，A8 < 0。
Suppose S9 > S5, A6+A7+A8+A9 > 0, that is 2 (A7+A8) > 0.
∵ A7 = 0，A8 < 0，∴ A7+A8 < 0。 The assumption is not true, so S9 < S5. C is wrong.
Answer: c
5. Let the series {an} be a geometric series, and the sum of the first n terms is Sn. If S3 = 3a3, the value of common ratio q is ().
A.- 12
C. 1 or-12 d.-2 or 12[
Analysis: let the first term be a 1 and the common ratio be q,
Then when Q = 1, S3 = 3A 1 = 3A3, which is suitable for the meaning of the question.
When q≠ 1, a1(1-Q3)1-q = 3a1Q2,
∴ 1-Q3 = 3Q2-3Q3, that is, 1+Q+Q2 = 3Q2, 2Q2-Q- 1 = 0,
The solution is q = 1 (truncation), or q =- 12.
To sum up, q = 1, or q =- 12.
Answer: c
6. If the general term formula of series {an} an = 5 252n-2-425n- 1, the largest term of series {an} is x term and the smallest term is y term, then x+y is equal to ().
a3 b . 4 c . 5d . 6
Analysis: an = 5252n-2-425n-1= 525n-1-252-45,
When n = 2, an is the smallest; When n = 1, an is the largest.
X = 1，y = 2，∴ x+y = 3。
A: A.
7. In the sequence {an}, a 1 = 15, 3an+ 1 = 3an-2 (n ∈ n *), then the product of two adjacent terms in the sequence is negative ().
a . a 2 1a 22 b . a22a 23 c . a23a 24d . a24a 25
Analysis: ∫3an+ 1 = 3an-2,
∴ an+ 1-an =-23, that is, the tolerance d =-23.
∴an=a 1+(n- 1)d= 15-23(n- 1).
Let an > 0, that is, 15-23 (n- 1) > 0, and the solution is n < 23.5.
And n∈N*, ∴n≤23, ∴ A23 > 0, and A24 < 0, ∴ A23A24 < 0.
Answer: c
8. The output value of a factory last year was A, and it is planned to increase by 10% every year in the next five years. From this year to the fifth year, the total output value of this factory is ().
a . 1. 14a b . 1. 15a
c . 1 1×( 1. 15- 1)a . d . 10×( 1. 16- 1)a
Analysis: Geometric series A 1 = A, W consists of known annual output value.
an = a( 1+ 10%)n- 1( 1≤n≤6)。
∴ the total output value is S6-a1=1× (1.15-1) a.
Answer: c
9. It is known that the sum of the first 20 items of arithmetic progression {an} composed of positive numbers is 100, so the maximum value of a7a 14 is ().
A.25b.50c. 100d。 Does not exist.
Analysis: From S20 = 100, A 1+A20 = 10. ∴ A7+A 14 = 10。
A7 > 0，A 14 > 0，∴ A7A 14 ≤ A7+A 1422 = 25。
A: A.
10. Let the sequence {an} be a geometric series with the first term m and the common ratio q(q≠0), and Sn is the sum of its first n terms. For any n∈N*, point an, S2nSn ().
A. MX+QY-Q = 0 on a straight line.
B. on a straight line, qx-my+m = 0.
C. qx+my-q = 0 on a straight line.
D. not necessarily in a straight line
Analysis: an = mqn- 1 = x, 1s2nsn = m (1-q2n)1-QM (1-qn)1-q =1+qn.
Substitute qn = y- 1 of ② into X = MQ (y- 1) of ①, that is, QX-my+m = 0.
Answer: b
1 1. Even series with 2 as the first item will be grouped as follows: (2), (4,6), (8, 10, 12), ... If there are n numbers in the nth group, then the first item in the nth group is ().
A.n2-n B.n2+n+2
C.n2+n D.n2-n+2
Analysis: Because the n- 1 group occupies the first1+2+3+…+(n-1) = (n-1) N2 item of the sequence 2, 6, …, the first item of the n-th group is the sequence 2.
Answer: d
12. Let the integer parts of m∈N* and log2m be represented by F(m), then the value of f (1)+f (2)+…+f (1024) is ().
A.8 204 B.8 192
C.9 2 18d。 None of the above is correct.
Analysis: according to the meaning of the question, f (1) = 0,
F (2) = F (3) = 1, there are two.
F (4) = F (5) = F (6) = F (7) = 2, there are 22.
F (8) = … = F (15) = 3, there are 23.
F (16) = … = F (3 1) = 4, there are 24.
…
F (512) = … = f (1023) = 9, and there are 29.
F (1 024) = 10, with 1.
Therefore, f (1)+f (2)+…+f (1024) = 0+1× 2+2× 22+3× 23+…+9× 29+10.
Let t = 1× 2+2× 22+3× 23+…+9× 29, ①
Then 2t =1× 22+2× 23+…+8× 29+9× 210. ②
①-②，get-t = 2+22+23+…+29-9×2 10 =
2( 1-29) 1-2-9×2 10=2 10-2-9×2 10=-8×2 10-2,
∴ t = 8× 210+2 = 8194m]
∴f( 1)+f(2)+…+f( 1 024)= 8 194+ 10 = 8 204。
A: A.
Volume 2 (non-multiple choice questions ***90 points)
Fill-in-the-blank question: This big question has four small questions, each with 5 points and * * 20 points.
13. If the series {an} satisfies the relation A 1 = 2 and AN+ 1 = 3an+2, the general term formula of the series is _ _ _ _ _ _ _ _.
Analysis: ∫an+ 1 = 3an+2 plus 1, an+ 1 = 3 (an+ 1),
∴ {an+ 1} is a geometric series with a 1+ 1 = 3 as the first term and 3 as the common ratio.
∴an+ 1=33n- 1=3n,∴an=3n- 1.
Answer: an = 3n- 1
14. In arithmetic progression {an} where the known tolerance is not zero, m = anan+3 and n = an+ 1an+2, then the relationship between m and n is _ _ _ _ _ _ _ _.
Analysis: Let the tolerance of {an} be d, then d≠0.
M-N=an(an+3d)-[(an+d)(an+2d)]
=an2+3dan-an2-3dan-2d2=-2d2