The sequence an is a decreasing proportional series, a2 a4 a6 21 8, a2a4a6 1 8, the sequence bn sati

1) In the case of proportional series, a2*a6=a4 squared. Then the cube of a4 is equal to 1 8, then a4 is equal to 1 2. Gotta :

a2 + a6 = 17 8, a2 * a6 = 1 4, these two equations are combined, eliminate a6, and get a2 = 1 8, or a2 = 2. When a2=1 8 and a4 > a2, then an is not a decreasing series, so a2=1 8 is rounded. Therefore:

a2=2，q=1/2，a1=4。then an=4 divided by (2 to the n-1 power). (2) Get an, and you can get it by satisfying the formula of that bn:

bn=3-n。Then the sum of the first n terms of bn is sn=(3-1)+(3-2)+(3-3)+3-(n-1))+3-n)=3n-(1+2+3+..

n) = (5n-n squared) 2

Summary. Proportional series refers to a series of numbers from the second term onwards, in which the ratio of each term to its predecessor is equal to the same constant, and is often represented by g and p. This constant is called the common ratio of the proportional series, which is usually represented by the letter q (q≠0), and the proportional series a1≠ 0.

Each of these items is not 0. Note: When q=1, an is a constant series.

The number column is proportional to the series a1+a2+a3+a4=8, a5+a6+a7+a8=16, find a13+a14+a15+a16

Extended Materials. The proportional sequence refers to a series of numbers from the second term onwards, in which the ratio of each term to its previous term is equal to the same constant, which is often represented by g and p. This constant is called the common ratio of the proportional series, which is usually represented by the letter q (q≠0), and the same is the ratio of the series a1≠ 0.

Each of these items is not 0. Note: When q=1, an is a constant series.

A n is equal to the n minus m power of am multiplied by q.

Can it be done with this formula.

You can also. How to do it with this formula.

Same. Write about the detailed process.

I'm above that.

Because a2a5=a3a4=32, a3 a4=12, and an is an increasing series, a3=4;a4=8, so q=2, a1=1

So an=2 n-1

Because bn-1 = 2bn 2an, (bn 1 an 1) = (bn an) 1

So it's a series of equal differences.

Solution: It is easy to find a1=, so an=10 (2-n)n is a natural number.

Let bn=logan=2-n

n is a natural number.

The sum of the first n terms is tn=2n-n(n+1) 2

loga1+loga2+..loga10=2*10-10*(10+1)/2=145

Let the stool travel ratio be q, then a3=2 jujube stool quietly void 3q, a5=2q 3, then 2 3q + 2q 3=20 9

Then an=2 3*3 (n-4)=2 3 (n-5) (n n+).

Then bn=log3 3 (n-5)=n-5, then b1=-4, b2=-3

So tn=-4+(-3)+(3) 2+(-3) 3+....+3) (n-1) is followed by a proportional sequence.

4+(-3)×[1-(-3)^(n-1)]/1-(-3)]

1/4*[19+(-3)^n]

Let un=an+1-an, u1=a2-a1=4-6=-2u2=a3-a2=3-4=-1

u2-u1=1

un equal difference series.

un=-2+(n-1)*1=n-3

an=a1+u1+u2+..un-1

a1+[-2+(n-4)](n-1)/2=6+(n-6)(n-1)/2

Let vn=bn-2

v1=b1-2=4

v2=b2-2=2

v3=b3-2=1

VN proportional series.

q=v2/v1=1/2

vn=4*(1/2)^(n-1)=(1/2)^(n-3)bn=(1/2)^(n-3)+2

an-bn=[6+(n-6)(n-1)/2]-[1/2)^(n-3)+2]

When n=1,2,3, an-bn=0

n=4,an=3,bn=2+1 2 an-bn=1 2n>4,an-an-1=un-1>1

bn-bn-1=(1/2)^(n-3)-(1/2)^(n-2)=(1/2)^(n-2)<1

an-bn>1+(an-1-bn-1)>1+(a4-b4)=1 2, so there is no k to make 0

<> is like imitating the picture, and the one who remembers it will give points, and I'm tired of it.

Solution: Let the common ratio of the series be q, and the first term is a1, then: an=(a1)q (n-1).

a2=(a1)q=2………1)a4=(a1)q^3=8………2)(2) (1), get: q 2=4

Solution: q1=2, q2=-2

Substituting (1), there is: (a1) (2) = 2

The solution is: (a1)1=1, (a1)2=-1

1. When a1=1, q=2

In this case, there is: an=2 (n-1), and the solution: a7=2 6=64

and: s10 = 1 (1-2 10) (1-2) = 10232, when a1 = -1, q = -2

In this case, there is: an=2 (n-1), and the solution: a7=2 6=64

and: s10 = (-1)[1-(-2) 10] [1-(-2)] = 1023 3.

The tolerance is da2*a6=a3 2

a1+d)*(a1+5d)=(a1+2d)^2d^2+2d*a1=0

d is not equal to 0d+2a1=0

d=-2a1

a1+a3+a5 sells a2+a4+a6

a1+a1+2d+a1+4d)/(a1+d+a1+3d+a1+5d)

3a1+6d)/(3a1+9d)

3a1-12a1)/(3a1-18a1)2.Open is auspicious, see is the fortune of the circle, may all the blessings flow to you, pray that you are in a good mood and all the best, may this good wish turn into the most sincere greetings to you!