Given that the line L1: a1x + b1y + 8 = 0 and L2: a2x + b2y + 8 = 0 intersect at one point (2, - 3), find the square of the line passing through two points (A1, B1) and (A2, B2) Come on Life saving Today's 23:00 deadline for bonus points

Given that the line L1: a1x + b1y + 8 = 0 and L2: a2x + b2y + 8 = 0 intersect at one point (2, - 3), find the square of the line passing through two points (A1, B1) and (A2, B2) Come on Life saving Today's 23:00 deadline for bonus points

The answer is... 2x-3y + 8 = 0

If a = (A1, A2, A3, A4) and B = (B1, B2, B3), what is the number of different mappings from a to B? (with explanation)

All numbers of a should have corresponding in B, and B may not be used up, so
Four pairs of 1:3 species; 3,1 pair of 2:4 * 3 * 2 = 24 species; 2,2 pair of 2:6 * 3 * 2 = 36; 2,1,1 pair of 3:6 * 3 * 2 = 36; 3 + 24 + 36 + 36 = 99 species

By the equation X4 + a1x3 + a2x2 + a3x + A4 = (x + 1) 4 + B1 (x + 1) 3 + B2 (x + 1) 2 + B3 (x + 1) + B4, define the mapping f: (A1, A2, A3, A4) → (B1, B2, B3, B4), then f (4, 3, 2, 1) equals ()
A. （1，2，3，4）B. （0，3，4，0）C. （-1，0，2，-2）D. （0，-3，4，-1）

Comparing the coefficients of X3 on both sides of the equation, we can get 4 = 4 + B1, then B1 = 0, so a and C are excluded; comparing the constant terms on both sides of the equation, we can find 1 = 1 + B1 + B2 + B3 + B4, ‖ B1 + B2 + B3 + B4 = 0. So B is excluded, so D should be selected

Let a = {(x, y) a1x + b1y + C1 = 0}, B = {(x, y) a2x + b2y + C2 = 0} find the solution set of two equations in the set, and express it with the expression related to a and B
Let the set a = {(x, y) - a1x + b1y + C1 = 0}, B = {(x, y) - a2x + b2y + C2 = 0} find the solution set of two equations in the set, and express the results with the expressions related to a and B
Find x, y

Solution set of two equations: a ∩ B

The first n terms of {an} and Sn satisfy Sn = 1-2 / 3an
Find LIM (a1s1 + a2s2 +...) AnSn).

When n = 1, S1 = 1 - (2 / 3) A1, A1 = 3 / 5
n> S (n-1) = 1 - (2 / 3) a (n-1)
An=Sn-S(n-1)=-(2/3)An+(2/3)A(n-1)
So an = (2 / 5) * a (n-1)
So {an} is an equal ratio sequence with a common ratio of 2 / 5
So an = (3 / 5) * (2 / 5) ^ (n-1)
So Sn = 1 - (2 / 3) an
=1-(2/5)*(2/5)^(n-1)
=1-(2/5)^n
AnSn=An*[1-(2/3)An]
=An-(2/3)An²
=An-(2/3)*(3/5)²*(2/5)^(2n-2)
=An-(6/25)*(4/25)^(n-1)
So a1s1 + a2s2 +... + ANSN
=(A1+A2+.+An)-(6/25)*[1+(4/25)+...+(4/25)^(n-1)]
=Sn-(6/25)*[1-(4/25)^n]/(1-4/25)
=1-(2/5)^n-2/7+(2/7)*(4/25)^n
=5/7-(2/5)^n-(4/25)^n
So LIM (a1s1 + a2s2 +) AnSn) (n→∞)
=5/7

The first n terms of sequence an and Sn satisfy Sn = 4 / 3an-1 / 3 * 2 ^ n + 1 + 2 / 3
, 1) find the first term A1 and the general term SN of the sequence (2) let the sum of the preceding terms of the sequence BN = 2 ^ n / Sn be TN to prove TN

(1) When n = 1, A1 = S1 = (4 / 3) A1 - (1 / 3) · 2 ^ (n + 1) + 2 / 3 - [(4 / 3) a (n-1) - (1 / 3) · 2 ^ (n + 1) + 4 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-1) · 2 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-1) + 2 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-1) + 4 ^ (n-

If the sum of the first n terms of the sequence {an} is Sn, Sn = 1-2 / 3an, then an=

sn+1-sn=an+1=-2/3(an+1-an),an+1=2an s1=a1,1-2/3a1=a1,a1=3/5.
a2=2a1=2 (3/5),an=2^(n-1)(3/5)

If the first n terms of the sequence {an} and Sn = 32An − 3, then the general term formula of the sequence is ()
A. an=2×3n-1B. an=3×2n-1C. an=2×3nD. an=3n+3

∵ Sn = 32An − 3, when n = 1, A1 = S1 = 32An − 3, then A1 = 6, when n ≥ 2, an = sn-sn-1 = 32An − 3 − & nbsp; (32An − 1 − 3) = 32 (an − an − 1) ∵ an = 3an-1, A1 = 6 ∵ sequence {an} is an equal ratio sequence with 6 as the first term and 3 as the common ratio ∵ an = 6.3n-1 = 2.3n, so select: C

The sum of the first n terms of the sequence an is Sn, 2An = Sn + 2, n ∈ n*
(1) The general term formula of the sequence an
(2) If BN = Log &; an + Log &; a (n + 1), find the first n terms and TN of the sequence {1 / BN (BN + 1)}

(1)2an =Sn+2n=1a1=22an =Sn+22[Sn-S(n-1)] =Sn+2Sn +2= 2[S(n-1)+2](Sn +2)/[S(n-1)+2] =2(Sn +2)/(S1 +2)=2^(n-1)Sn +2 =2^(n+1)Sn = -2+ 2^(n+1)an =Sn-S(n-1) = 2^n(2)bn = logan + loga(n+1)= n+(n+1)= 2n+11/[...

In the sequence {an}, if A1 = 1,3anan-1 (Note: n-1 is the whole) + an-an-1 (Note: n-1 is the whole) = 0 (n is greater than or equal to 2, n belongs to n), find the general term an

Let BN = 1 / an, multiply the original equation by 1 / (ANA (n-1)), then bn-b (n-1) = 3, BN = B2 + (n-2) * 3, (n > = 2) and B1 = 1, then an = 1 / (3n-2)