It is known that n is a positive integer, PN (xn, yn) is an inverse scaling function y = KX, where X1 = 1, X2 = 2 , xn = n, T 1 = x 1y 2, T 2 = x 2Y 3 If T1 = 1, then T1 ·t2 · The value of T9 is______ ．

It is known that n is a positive integer, PN (xn, yn) is an inverse scaling function y = KX, where X1 = 1, X2 = 2 , xn = n, T 1 = x 1y 2, T 2 = x 2Y 3 If T1 = 1, then T1 ·t2 · The value of T9 is______ ．

T1•T2•… •Tn=x1y2•x2y3… xnyn+1=x1•kx2•x2•kx3•x3•kx4… Xn · kxn + 1 = x1 · knxn + 1, and because X1 = 1, n = 9, and because T1 = 1, so x1y2 = 1, and because X1 = 1, so y2 = 1, that is kx2 = 1, and X2 = 2, k = 2, so the original formula = 29x9 + 1, then T1 · T2 · •T9=x1（y2•x2）（y3•x3）… (Y9 · x9) Y10 = 29x9 + 1 = 2910 = 51.2

(2013 · Nankai District simulation) let the sum of the first n terms of the equal ratio sequence {an} be Sn, if S10: S5 = 1:2, then S15: S5 is equal to ()
A. 3：4B. 2：3C. 1：2D. 1：3

In the equal ratio sequence, the sum of every five items is still equal ratio sequence. Suppose S5 = x, then from the condition, we can get S10 = 12x, ｜ s10-s5 = 12 & nbsp; x-x = - 12 & nbsp; X, ｜ s15-s10 = 14 & nbsp; X, ｜ S15 = 12 & nbsp; X + 14x = 34x, so S15: S5 = 34xx = 34, so select a

Let Sn denote the sum of the first n terms of the equal ratio sequence {an} (n ∈ n *). If s10s5 = 3, then s15s5 = 3=______ ．

∵ {an} is an equal ratio sequence, let the common ratio be q, when q = 1, s10s5 = 2 does not conform to the meaning of the problem; when Q ≠ 1, s10s5 = A1 (1 − Q10) 1 − QA1 (1 − Q5) 1 − q = 1 − Q101 − Q5 = 3 (Q ≠ 1), {q10-3q5 + 2 = 0, the solution is Q5 = 1 (rounding off) or Q5 = 2.} s15s5 = 1 − q151 − Q5 = 7

It is known that the first term of the arithmetic sequence {an} is A1 = 20, and the sum of the first n terms is Sn, which satisfies S10 = S15. When we find out the value of N, we can get the maximum value of Sn and get the maximum value

∵ A1 = 20, S10 = S15, ∵ 10 × 20 + 10 × 92d = 15 × 20 + 15 × 142d, d = - 53 (3 points) the number sequence is decreasing. The general term formula is an = - 53n + 653. A13 = 0 When n = 12 or n = 13, the maximum value of Sn is S12 = S13 = 130 (12 points)

It is known that the first term of the arithmetic sequence {an} is A1 = 20, and the sum of the first n terms is Sn, which satisfies S10 = S15. When we find out the value of N, we can get the maximum value of Sn and get the maximum value

∵ A1 = 20, S10 = S15, ∵ 10 × 20 + 10 × 92d = 15 × 20 + 15 × 142d, d = - 53 (3 points) the number sequence is decreasing. The general term formula is an = - 53n + 653. A13 = 0 When n = 12 or n = 13, the maximum value of Sn is S12 = S13 = 130 (12 points)

It is known that the first term of a sequence an is an arithmetic sequence of 1, and an + 1 > an, if A3, a7 +, 3a9 are equal ratio sequence
Finding the general term of an formula
Let the sum of the first n terms of an be Sn, f (n) = Sn / (n + 18) Sn + 1. When we ask what the value is, f (n) is the largest, and we find the largest value

An = 1 + (n-1) DA (n + 1) > an, that is 1 + nd > 1 + (n-1) d, so d > 0
a3=1+2d
a7+2=1+6d+2=3+6d
3a9=3(1+8d)
(a7+2)^2=a3×3a9
(3+6d)^2=(1+2d)×3(1+8d)
The solution of the equation is d = 1 or D = - 1 / 2
Because d > 0, d = - 1 / 2 is omitted
Then an = a1 + (n-1) = 1 + (n-1) * 1 = n
Sn=n(n+1)/2
f(n)=[n(n+1)/2]/[(n+18)*(n+1)(n+2)/2]
=n/(n+18)(n+2)
=n(n^2+20n+36)
=1 / (n + 20 + 36 / N) (because n + 36 / N > = 2 * √ (n * 36 / N) = 12)

If LiMn → + ∞ Sn + 1sn = 1, then the value range of common ratio q is ()
A. q≥1B. 0＜q＜1C. 0＜q≤1D. q＞1

When q = 1, Sn + 1 = (n + 1) A1, so LiMn → + ∞ Sn + 1sn = n + 1n = 1 holds, when Q ≠ 1 is yes, Sn = & nbsp; A1 (1 − QN) 1 − Q, so LiMn → + ∞ Sn + 1sn = 1 − QN + 11 − QN, we can see that LiMn → + ∞ Sn + 1sn = 1 holds when q is a fraction less than 1, so the answer should choose C

If we know that {an} is an infinitely proportional sequence and lim (SN) = 1 / 2, then the range of A1 is? Online, etc!
If we know that {an} is an infinitely proportional sequence and lim (SN) = 1 / 2, then the range of A1 is?
Wait online!

a1/(1-q)=1/2
a1=1/2·(1-q)
[where, - 1 < Q < 1]
∴0＜a1＜1

It is proved that if a bounded real number sequence has only one accumulation point, then the sequence converges and the limit = accumulation point

If the limit does not exist, then there is infinite x (NK) outside a neighborhood s (x, ε) of an accumulation point X of X (n). Because the sequence x (NK) is also a bounded sequence, there is also an accumulation point y, and | Y-X | ≥ ε > 0. This is contrary to the condition

When to use Chebyshev inequality and central limit theorem
I can't make it clear

Their conditions are different
Chebyshev is more relaxed, as long as ξ 1, ξ 2 D ξ K is uniformly bounded, but the result is only one
It's qualitative,
The central limit theorem is much stronger Apart from being independent of each other, they should have the same distribution
And so on. There are several theorems, conditions are different, and the results are qualitative and more quantitative
When using, as long as the conditions are good, try to use the central limit theorem
There are many inequalities