∴ 1/an=[(a(n- 1)+2n-2]/nb*a(n- 1)= 1/nb+2(n- 1)/nba(n- 1)

∴nb/an=2(n- 1)/a(n- 1)+ 1

Let n/an=cn

bcn=2c(n- 1)+ 1

∴cn=2/bc(n- 1)+ 1/b= 1/b[2c(n- 1)+ 1]

∴cn+ 1/(2-b)=2/b[c(n- 1)+ 1/(2-b)]

∴ {cn+1(2-b)} is c1+1(2-b) =1/a1+1(2-)

∴cn+ 1/(2-b)=2/b(2-b)*(2/b)^(n- 1)=(2-b)*(2/b)^n

∴cn=)=(2-b)*(2/b)^n- 1/(2-b)=[(2-b)^2*2^n-b^n]/(2-b)*b^n

That is, n/an = [(2-b) 2 * 2 n-b n]/(2-b) * b n.

∴an=n*(2-b)*b^n/[(2-b)^2*2^n-b^n]

The second question is proved by mathematical induction.

This is mainly used in the construction of sequences, substitution methods, formulas and so on. I hope it helps you!