It is proved that if the first term of geometric series is a 1 and the common ratio is q, then:
AK = a 1 q^(k- 1),al=a 1 q^(l- 1),am=a 1 q^(m- 1),an=a 1 q^(n- 1)
So:
ak*al=a^2*q^(k+l-2),am*an=a^2*q(m+n-2)，
Therefore: ak*al=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) a(n-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)+a (n-k) = a1+an In arithmetic progression, A4+A6+A8+A10+A12 =120, then 2A9.
a20 b . 22 c . 24 D28
Solution: From a4+a 12=2a8, a6+a 10 =2a8, given conditions:
5a8= 120，a8=24
And 2a9-a10 = 2 (a1+8d)-(a1+9d) = a1+7d = A8 = 24.
So choose C, let Sn be the sum of the first n terms of arithmetic progression, S9= 18, a (n-4) = 30 (n >; 9), Sn=336, then n is ()
2 1 C.9 D.8
Solution: Since S9=9×a5= 18, a5=2, a5+a(n-4)=a 1+an=2+30=32, therefore, n=2 1, let arithmetic progression satisfy 3a8 = 5a6544. 0, and Sn is the sum of the first n items, 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, a1=-19.5d.
Make an ≥ 0 → n ≤ 20; When n >; At 20 o'clock, Ann < 0
∴S 19=S20 maximum, select (c)
Note: Positive odd sets {1, 3,5, ...} can also be grouped by (2n- 1) odd numbers according to quadratic function from small to large according to the nth group:
{ 1}, {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) = n 2 block.
And1991= 2× 996-1is the 996th positive odd number.
∵3 1^2=96 1<； 996 & lt 1024=32^2
∴ 199 1 should belong to group 3 1+ 1=32.
So fill in 32 known sets M={x, xy, lg(xy)} and N={0,∣x∣, y}
And M=N, then
(x+ 1/y)+(x^2+ 1/y^2)+(x^3+ 1/y^3)+？ The value of+(x2001+1/y2001) 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. It is known that the sequence satisfies 3a(n+ 1)+an=4(n≥ 1) and a 1=9, and the sum of the first n terms is Sn, then the inequality ∣ Sn-n-6 ∣.
The smallest integer n of is ()
5 (B)6 (C)7 (D)8
1994 national high school mathematics league exam questions
Solution: For 3a(n+ 1)+an=4:
3[a(n+ 1)- 1]=-(an- 1)
a(n+ 1)/an=- 1/3
an=8*(- 1/3)^(n- 1)+ 1
sn=8{ 1+(- 1/3)+(- 1/3)^2+……+(- 1/3)^(n- 1)]+n
=6-6*(- 1/3)^n+n
|sn-n-6|=|-6*(- 1/3)^n|<； 1/ 125
When n=7, it meets the requirements, so choose C.
Note: The series is neither arithmetic progression nor geometric progression, but a series consisting of the sum of two corresponding items in arithmetic progression: 1, 1, …, 1 and geometric progression: 8, -24, 72, -2 16, …, and the items are equal, so the first n. Let the sum of the first n terms of the sequence Sn=2an- 1(n= 1, 2, ...), and the sequence satisfies b 1 = 3, b (k+ 1) = AK+BK (k = 1, 1996 the first question in the second exam of the national senior high school mathematics league.
Solution: From Sn=2an- 1, let n= 1, and get S1= a1= 2a1-1,∴ a1=
And Sn=2an- 1 ②.
s(n- 1)= 2(an- 1)- 1③
②-③De:Sn-S(n- 1)= 2an-2a(n- 1)
∴an=2an-2a(n- 1)
So an=2a(n- 1)
∴ series is a geometric series with a 1= 1 as the first term and q=2 as the common ratio, so an = 2 (n- 1) ④.
By ⑤ b (k+1) = AK+bk (k =1,2, ...),
∴ Add the above formulas to get {bn-2}={an},
∴{bn}=2^(n- 1)+2
∴Sbn=2^n- 1+2*n