The sequence an satisfies a1 1 a2 3 2 an 2 3 2an 1 1 2an so that bn 3n 2 is calculated

Because an+2=3 2an+1-1 2an

So an+2-an+1=1 2an+1-1 2an=1 2(an+1-an).

And because a2-a1=3 2-1=1 2

Therefore, the number column is a proportional series with the first term 1 2 and the common ratio is 1 2.

i.e. an+1-an=(1 2) n

an+1-a1=(1/2)^n+(1/2)^(n-1)+…1/2=1-(1/2)^n

So an+1=2-(1 2) n

an=2-(1/2)^(n-1)

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

sn=(1+4+……3n-2))-1*1+4*(1/2)+…3n-2)*(1/2)^(n-1)]

Let tn=[1*1+4*(1 2)+....3n-2)*(1/2)^(n-1)]

It can be seen that 2tn-tn=[1*2+4*1+......3n-2)*(1/2)^(n-2)]-1*1+4*(1/2)+…3n-2)*(1/2)^(n-1)]

2+3[1+(1/2)+…1/2)^(n-2)]-3n-2)*(1/2)^(n-1)

8-3*(1/2)^(n-2)-(3n-2)*(1/2)^(n-1)

8-(3n-8)*(1/2)^(n-1)

And because, 1+4+......3n-2)=n(3n-1)/2

So, sn=n(3n-1) 2-tn=n(3n-1) 2-8+(3n-8)*(1 2) (n-1).

an+2=3 2an+1-1 2an is not clearly described, you can probably guess it.

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

a(n+1)=a(n)+4/3;Arithmetic progression.

But is n starting from ** n>=2?

If so, it is a(n)=3 2+(n-2)*4 3=4n 3-7 6a(n)-b(n)=-5n 3-19 6

s(n) ask for it yourself, n>=2

is 1 2an=1 2a(n-1)+1, multiply both sides by 2 at the same time to get 1 an=1 a(n-1)+2 then (1 an) can be regarded as an equal difference series a1=1 from the second push spike nucleus 1 an=2n-1 an=1 (2n-1) a1*a2+a2*a3+...an*an+1=1/1*3+1/3*5+..1/(2n-1)*(2n+1)>16/33 1/1*3+1/3*5+..

1 (2n-1)*(2n+1) is obtained by the elimination of the rift term = (1-1 collapse (2n+1)) 2

1-1 (2n+1)) 2 >16 33 Launched Trapped Circle N>16

an+2 - an+1=1/2an+1 - 1/2an

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

The number sequence {an - an-1} is a proportional series with a common ratio of 1 2.

an - an-1=(a2-a1)1/2^(n-2)=1/2^(n-1)

So we get an-1 - an-2=1 2 (n-2).

a3-a2=1/2^2

a2-a1=1 2 Add left and left, and right and right add to get.

an-a1=1/2^(n-1)+1/2(n-2)+.1/2^2+1/2

an=1/2^(n-1)+1/2(n-2)+.1/2^2+1/2+1/2^0

an=(1-1/2^n)/(1-1/2)

2-1/2(n-1)

When n=1, a1=2-1=1

When n=2, a2=2-1 2=3 2

So an=2-1 2(n-1) (n=>1, n is a positive integer).

a1a2+a2a3+…+ana(n+1)=na1a(n+1)

a1a2+a2a3+…+a(n-1)ana=(n-1)a1an

Subtract the two formulas.

ana(n+1)=na1a(n+1)-(n-1)a1an

Both sides of the equation are divided by ana(n+1) at the same time

1=na1/an-(n-1)a1/a(n+1)

1/a1=n/an-(n-1)/a(n+1)

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

n/an-(n-1)/a(n+1)=4...1

In the same way, (n-1) a(n-1)-(n-2) an=4....2

1-2 formula.

2(n-1)/an-(n-1)/a(n+1)-(n-1)/a(n-1)=0

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

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

So 1 an is a series of equal differences

d=1/a2-1/a1

11/an=1/a1+(n-1)d

1/(1/4)+n-1

4+n-1n+3

1/a1+1/a2+..1/a97