In the known sequence An, Sn 1 4An 2, A1 1, let Bn An 1 2An, and verify that Bn is an equal proporti

s(n+1)=4an+2

sn=4a(n-1)+2

Subtracting the two formulas, we get a(n+1)=4an-4a(n-1) and a(n+1)-2an=2(an-a(n-1)) is a proportional series with a common ratio of 2.

The following are the general items, and I want to see them.

and s2=4a1+2=6, s2=a1+a2, so a2=5, so a(n+1)-2an=3*2 (n-1) divide by 2 (n+1) on both sides of the equation, and get a(n+1) 2 (n+1)-an 2 n=3 4

So it is an equal difference series, the first term a1 2=1 2, and the tolerance is 3 4an 2 n=1 2+3 4(n-1)=(3n-1) 4an=(3n-1)2 (n-2).

s(n+1)=4a(n)+2, 4a(1)+2=4+2=s(1+1)=a(1)+a(2)=1+a(2), a(2)=5.

s(n+2)=4a(n+1)+2,a(n+2)=s(n+2)-s(n+1)=4a(n+1)-4a(n),a(n+2)-2a(n+1)=2[a(n+1)-2a(n)].

a(n+1)-2a(n)} is the first proportional sequence with a(2)-2a(1)=5-2=3 and a common ratio of 2.

a(n+1)-2a(n)=3*2 (n-1),b(n)=a(n+1)-2a(n)=3*2 (n-1),b(n)=a(n+1)-2a(n)} is the first proportional sequence with 3 terms and 2 common ratios.

When n 2, sn=4a(n 1) 2, subtract it from s(n 1)=4an 2, and obtain: a(n 1) = 4an 4a(n 1), i.e., a(n 1) 2an=2[an a(n 1)], then:

bn=2b(n 1), where n, then a2=3a1 2=5, b1=a2 2a1=3It is a proportional series with 3 as the first term per mu and q=2 as the common ratio, and obtains: bn=3 2 (n 1)

That is: a(n 1) 2an=3 2 (n 1), the two sides are divided by 2(n 1), and the results are: [a(n 1)] 2 (n 1)] an] [2 (n 2)]=3=constant, then the series is a series of equal differences with a1 [2 (1 2)]=2 as the first term and d=3 as the tolerance, then an [2 (n 2)]=2 3(n 1)=3n 1, therefore, an=(3n Xunqingzi1) 2 (n 2)

a(n+1)=s(n+1)-s(n)=4a(n+1-1)+1-(4a(n-1)+1)=4a(n)-4a(n-1)=2a(n)+2a(n)-4a(n-1)a(n+1)-2a(n)=2(a(n)-2a(n-1))b(n)=a(n+1)-2a(n)b(n)=2b(n-1)s2=4a1+1=5a2=s2-a1=4b1=a2-2a1=2 is not 0bn for the proportional series, the first Feng Shen quiet term is 2, the male blind ruler ratio is silver slag 2, b (..

bn=(3an-2)/(an-1)

an=(bn-2)/(bn-3)

a(n+1)=[b(n+1)-2]/[b(n+1)-3]

a(n+1)=(4an-2)/(3an-1)

3a(n+1)an-a(n+1)=4an-2

3[(bn-2)/(bn-3)]-b(n+1)-2]/[b(n+1)-3]=4(bn-2)/(bn-3)-2

3[b(n+1)-2](bn-2)-[b(n+1)-2](bn-3)=4(bn-2)[b(n+1)-3]-2[b(n+1)-3](bn-3)

2b(n+1)bn-3b(n+1)-4bn+6=2b(n+1)bn-2b(n+1)-6bn+6

b(n+1)=2bn

bn=b1*2^(n-1)

b1=(3a1-2)/(a1-1)=4

bn=b1*2 (n-1)=2 (n+1) is the ratio of hunger and wisdom;

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

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

3-2^(n+1)]an=2-2^(n+1)

an=[2-2 (n+1)] 3-2 (n+1)],1,Slow in known sequence,a1=2,an+1=4an-2 3an-1 bn=3an-2 an-1 verify; A sequence is a proportional sequence and a general formula.

∵s(n+1)=4an+2

When n 2, sn=4a(n-1)+2

s(n+1)-sn=4an-4a(n-1), i.e.: a(n+1)=4an-4a(n-1).1)

a(n+1)-2an=2[an-2a(n-1)], i.e., bn=2b(n-1)

It's a proportional series.

The common ratio of the proportional series is 2

The first term b1 a2-2a1, and s2 4a1+2, a1+a2=4a1+2, a2=3a1+2=5, b1=3

The general formula for a series of numbers is: bn=3*2 (n-1)

is obtained by a1=, s2=4a1+2=6=a1+a2, so a2=5 is obtained by (1) and the common ratio is 2, and the first term is a2-2a1=3 proportional series, so a(n+1)-2an=3*2 (n-1) is divided by 2 (n+1) on both sides.

a(n+1) [2 (n+1)]-an 2 n=3 4, so the sequence an 2 n is a series of equal differences. (Tolerance is 3 4)an 2 n=1 2+3 4(n-1)=3 4n-1 4an=3*[2 (n-2)]*n-2 (n-2) =(3n-1)*2 (n-2).

cn=an/(3n-1)=（3n-1）*2^(n-2)/(3n-1)=2^(n-2)

So the sequence is a series of numbers with the first term 1 2 and the common ratio is 2.

From a1=1, and sn+1=4an+1, we get.

a1+a2=4an+1, a2=3a1+1=4, b1=a2-2a1=2, by sn+1=4an+1....①

Then when n 2, there is sn=4an-1+1....②

an+1=4an-4an-1, an+1-2an=2(an-2an-1).

and bn=an+1-2an bn=2bn-1 is the first proportional series of terms b1=2 and the common ratio is equal to 2

The proposition is not true, when n=1, sn+1=4an+1 can be reduced to a1+1=4a1+1, since a1=1 1 is not equal to 4+1So the proposition is not valid, I don't know if you made a mistake, the brother upstairs, did you make it to test.

b(n+1)=[a(n+1)-2]/[a(n+1)+1]=[(3an+2)/(an+2)-2]/[3an+2)/(an+2)+1]

an-2/4an+4

bn=an-2/an+1

Therefore, bn+1 bn=1 4

So it's a proportional series.

It is known that the sequence an is adjacent to two terms, an, and an 1 is the two real roots of the equation x 2-(2 n)*x bn=0, and a-1 3 is a proportional sequence with a common ratio of -1 when n is even. And the general formula for an is an =

You can substitute the formula of an+1 into bn as a whole.

a1=1,s(n+1)=4an+2, please enclose the parentheses when you copy the question, otherwise you will misunderstand and the question will be wrong.

In the case of n=2, a1 + a2 =4a2 +2 gives a2=5, and also makes n=3, giving a3=16

Use a(n+2)=s(n+2)-s(n+1)=4a(n+1) -4an

Shift the term to obtain a(n+2)-2a(n+1)=2a(n+1) -4an

We get [a(n+2)-2a(n+1)] a(n+1) -2an] =2

i.e. a(n+1)-2an is proportional. Let bn=a(n+1) -2an, then bn is a proportional series.

b2 b1=(a3 -2a2) (a2 -2a1)=2, the common ratio is, by the proportional series bn=3*2 (n-1), that is, a(n+1) -2an=3*2 (n-1).

A(n+1)= 2an+3*2 (n-1) is iterated sequentially.

a(n+1)= 2an+3*2^(n-1)=2[2a(n-1) +3*2^(n-2)]+3*2^(n-1)=2^2 *a(n-1)+2*3*2^(n-1)

2^3 *a(n-2)+3*3*2^(n-1)=.2^n *a1 +n *3*2^(n-1)=a1*2^n +n3*2^(n-1)

2^n +n3*2^(n-1)

an=2^(n-1) +n-1)3*2^(n-2)=2^(n-2) *3n-1)。So an (3n-1) = 2 (n-2).

So when cn=an (3n-1), cn is equivalent.

s(n+1)-sn=4(an-a(n-1)), i.e. a(n+1)=4(an-a(n-1)).

b(n+1)=a(n+1)-2an=2(an-2a(n-1))=2bn

Since you've already asked the first question, I'll skip it.

s2=4a1+2

a1+a2=4a1+2

a2=3a1+2=5

b1=a2-2a1=5-2=3

bn=3*2^(n-1)

a(n+1)-2an=3*2^(n-1)

Divide by 2 (n+1) on both sides at the same time

a(n+1) 2 (n+1)-an 2 n=3 4 so into an equal difference series, with a tolerance of 3 4

an/2^n=1/2+3/4(n-1)=3/4n-1/4an=3*[2^(n-2)]*n-2^(n-2)