s[ 1]= a[ 1]So(a[ 1]- 1)？ =a[ 1]S[ 1] ==》a？ [ 1]-2a[ 1]+ 1 = a？ [ 1]-& gt； a[ 1]= 1/2

(S[2]- 1)？ =a[2]S[2] (a[2]- 1/2)？ =a[2](a[2]+ 1/2)

Answer? [2]-a[2]+ 1/4=a？ [2]+( 1/2)a[2] 1/4 =(3/2)a[2]-& gt； a[2]= 1/6 S[2]=2/3

(S[3]- 1)？ =a[3]S[3] (a[3]- 1/3)？ =a[3](a[3]+2/3)

Answer? [3]-(2/3)a[3]+ 1/9=a？ [3]+(2/3)a[3] 1/9 =(4/3)a[3]-& gt； a[3]= 1/ 12 S[3]=3/4

The law s [n] = n/(n+1) a [n] =1[n (n+1)] =1(n+1).

Actually (S[n]- 1)? = a[n]S[n]=(S[n]-S[n- 1])S[n]

s？ [n]-2S[n]+ 1=S？ [n]-S[n- 1]S[n]

1=(2-S[n- 1])S[n]

S[n]= 1/(2-S[n- 1])

1/( 1-S[n])=(2-S[n- 1])/( 1-S[n- 1])= 1/( 1-S[n- 1])+ 1

1(1-s [1]) = 2 so1(1-s [n]) = n+1.

S[n]= 1- 1/(n+ 1)= n/(n+ 1)a[n]= S[n- 1]= 1- 1/(n+ 1)-[ 1- 1/n]= 1/n-654338

a[20 15]+ 1/20 16 = 1/20 15- 1/20 16+ 1/20 16 = 1/20 15