How do you solve a system of equations in a proportional series?

You should ask the first question, this does not need to solve the specific value, solve the expression form of cn about an, the ratio is a fixed value to prove it, <>

<> figure limbs to cut off the posture to solve this mess.

This is a system of binary equations, because it is not a system of binary equations, but a system of binary higher-order equations, and the common ratio of the proportional series may be zero, so we can divide the two equations together, which is called multiplication and division.

Divide the two equations, divide the second equation by the first equation, get q = 8, use the direct opening method, find q = 2, indicating that the common ratio of this proportional series is 2 (the common ratio of the proportional series cannot be an imaginary number), substitute the first infiltration group equation, 2a1=10, a1=5, that is, Cong Beicheng said that the common ratio of this proportional series is 2, and the first term is 5.

a1q=10 --

a1p^4=80---

Honor: q 3=8, q = 2

Dai La trembling into the old game of caution: a1 = 5

Equation below: a1(q q 4)=42

Divide the two formulas and subtract a1:

1+q²+q³)/q－q^4)=4

Then 1+q +q =4 (q q 4).

1+q²+q³－4q+4q^4=0

Then Zao Kai then solves the equation to calculate q. But the value of your Q is very complicated, and it is unlikely to be with Bi. What is the original question?

This well-known problem can only be solved numerically. Divide the two equations to give the 4th order equation. The numerical solution of the book of virtual skin obtains:

q1=q2=

a11=a12=

an >0

an=a1+a3=10

a1.(1+q^2) =10 (1)

a2+a3=6

1+q 2) [q(1+q)] 5-year-old model 33(1+q 2) =5q(1+q).

2q^2+5q-3=0

2q-1)(q+3)=0

q=1/2 or -3(rej)

from (1)

a1.(1+q^2) =10

a1.(1+1/4) =10

a1=8an = 8.(1/2)^(n-1)bn log<2>an

log<2> [8.(1/2)^(n-1) ]log<2> 2^(4-n)

4-n> is the number of equal-difference fibrillary dates.

snb1+b2+..bn

n( 3+4-n)/2

n(7-n)/2

It's okay to subtract it one step at a time.

1) Which column of the proportional beam: a(n+1) bending slag guess an=q, n is a positive integer.

2) General formula: an=a1 q (n 1);

Promotional: an=am q (n m);

3) Find the buried type and formula: sn=n*a1(q=1)sn=a1(1-q n) (1-q).

a1-anq) (1-q)q is not equal to.

4) Nature: If.

m, n, p, q n, and m n=p q, then am·an=ap*aq;

In the proportional series, each in turn.

The sum of k terms is still an equal proportional sequence.

5) "g is the proportional middle term of a and b", "g 2=ab(g≠0)".

6) In the proportional series, neither the first term A1 nor the common ratio q are zero.

Note: In the above formula, a n denotes a to the nth power.

The first equation is squared on both sides:

a1²q^6+a1²q^12+2a1²q^9=4 ..3)

3) (2) De:

1/q³+q³+2=-1/2

1/q³+q³+5/2=0

Multiply both sides by 2q:

2+2q^6+5q³=0

2q³+1)(q³+2)=0

q = -1 2, or q = -2

q 9 = -1 8, or q 9 = -8

Substituting q 9 = -1 8 into a1 q 9 = -8 yields:

a1²=64

a1 = -8 or 8

Substituting q 9 = -8 into a1 q 9 = -8 yields:

a1 = 1a1 = -1 or 1

The first equation is approximately posted.

That is, 1-q 4+1-q = 2-2q

q is obviously not equal to 0

So i.e. q + q-2 = 0

Obviously, q is not equal to 1

So q=-2

Substitute the second equation to find a3.

2 formula 1 formula, get q 4 = 9 16, solve q1 = 3 or q2 = -3 , substitute q1 or q2 into 1 formula, get a = 4 9.