[S(n+ 1)-Sn]/[Sn-S(n- 1)]= q

* 1

[a (n+1)+1]/[an+1] = k (constant)

[s (n+1)-sn+1]/[sn-s (n-1)+1] = k (constant)

*2

Substitute * 1 into * 2.

free

k = q+( 1-q)/[Sn-S(n- 1)+ 1]

K.q is a constant value.

So Sn-S(n- 1)

For colonization

Ann is colonization.

Is a constant sequence.

Sn=2n

2. The square of the proportional term is equal to the product of two terms.

This is already a good idea.

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

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

Tn= 1/2+3/(2^2)+5/(2^3)+...+(2n- 1)/(2^n)+k

@ 1

( 1/2)tn= 1/(2^2)+3/(2^3)+...+(2n-3)/(2^n)++(2n- 1)/[2^(n+ 1)]+0.5k

@2

@ 1-

@2

Get Tn