It is known that between the general formula an of the sequence of an and the formula of the first n

1) From sn=2-3an, a1=s1=2-3*a1, so a1=1 2 is also because an=sn-s(n-1)=2-3an-(2-3a(n-1))=3a(n-1)-3an

So 4an = 3a(n-1).

So it is a proportional series, and the common ratio q=3 4

Therefore, the general term an=(1 2)*[3 4) to the power of (n-1)]2) is brought into the equation for summing the proportional series by the result of 1).

sn=2 [1-(3, 4) to the nth power].

Solution: When n=1, s1=a1=2-3a1

4a1=2a1=1/2

At n 2, sn=2-3an

s(n-1)=2-3a(n-1)

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

an a(n-1)=3 4, is a fixed value.

The number series is an equal proportional series with 1 2 as the first term and 3 4 as the common ratio.

an=(1/2)×(3/4)^(n-1)

The general formula for the series is: an=(1, 2), (3, 4) (n-1), and the answer is correct.

Analyze your thoughts:

First of all, the categorical discussion is right.

At n 2, s(n-1) is actually s1 and therefore includes the case when n = 1. The reason why we need to classify the discussion and separate the cases of n 2 is just to make s(n-1) meaningful.

sn=2-3an

sn-1=2-3an-1

sn-sn-1=an

an=3an-1-(3an)

4an=3an-1

an/(an-1)=3/4

A number column is a proportional series with a common ratio of 3 to 4.

2) Find A1 and Q according to the general terms

a(1)=s(1)=2-3a(1),a(1)s(n)

3a(n),s(n+1)

2-3a(n+1),a(n+1)=s(n+1)-s(n)3a(n)3a(n+1),a(n+1)

3/4)a(n),a(n)}

It is a proportional series with the first term a(1) = 1 2 and the common ratio (3 4).

a(n)1/2)(3/4)^(n-1)

s(n+1)

3a(n+1)

3[s(n+1)-s(n)],4s(n+1)3s(n)2,s(n+1)

3/4)s(n)

sn=2an-3n

s(n-1)=2a(n-1)-3(n-1) subtract from the two pies.

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

Therefore, an is a series of numbers (an-3) (an-1)-3)=2

Solution: when n=1, a1=s1=2a1+1, i.e., a1=-1, when n>=2, sn=2an+n(n-1)=2a(n-1)+(n-1) subtract the two formulas to obtain: an=2an-2a(n-1)+1, that is, an=2a(n-1)-1, that is, an-1=2[a(n-1)-1] is the same ratio of the feast a1-1=-2 and 2 is the common ratio: an-1=-2(2) (n-1)=-2 n an=1-2 ..

a1=s1=3a1-2, so a1=1, and sn=3an-2ns(n-1)=3a(n-1)-2(n-1), so an=sn-s(n-1)=3an-3(n-1)-2, i.e., an=3 2a(n-1)+1, so an+2=3 2(a(n-1)+2) is the first term a1+2=3, and the common ratio is 3 2 proportional series an+2=3*(3 2) (n-1)=2*(3 2) n, so an=2*(3 2). )..

a(n)=s(n)-s(n-1)=3^n-2^na(n-1)=s(n-1)-s(n-2)=3^n-1-2^n-1...Recursively a(2)=s(2)-s(1)=3 2-2 2a(1)=s(1)=3-2=1 Add the left and right epoch clusters of all the above formulas to get limb cherry a(1)+a(n)=s(n)=Actually, it is the two that are 3 and 2.

a1＝-1a2＝-4

A3 -7 equal difference series, the tolerance is d 3

an -1 + (n-1) -3 (general formula).

sn=(a1+an)n/2

sn＝（an-1）n/2

sn=2an-3n

s(n-1)=2a(n-1)-3(n-1).

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

So an is a series of equal differences (an-3) ((an-1)-3)=2

Solution: Let n=1

a1=s1=(1/2)(1-a1)

I tidy it up and get it. 3a1=1

a1=1/3

sn=(1/2)(1-an)

sn-1=(1/2)[1-a(n-1)]

an=sn-sn-1=(1/2)(1-an)-(1/2)[1-a(n-1)]

I tidy it up and get it. 3an=a(n-1)

an (an-1)=1 3, which is a fixed value.

A number series is a proportional series with 1 3 as the first term and 1 3 as the common ratio.

The general formula for the series of an=(1 3)(1 3) (n-1)=1 3 is an=1 3.