a(n+ 1)=2an+3？

a(n+ 1)-3^(n+ 1)=2an+3？ -3^(n+ 1)=2an+3？ -3×3? =2an -2×3？ =2 (An -3? )

[a(n+ 1)-3^(n+ 1)]/(an-3？ ) =2, which is a fixed value.

a 1 -3= 1-3=-2

Sequence {An -3? } is a geometric series with -2 as the first term and 2 as the common ratio.

An -3? =(-2)×2^(n- 1)=-2？

an=3？ -2?

When n= 1, a 1=3-2= 1, which is also satisfied.

The general formula of the sequence {an} is an=3? -2? .

2.

a(n+ 1)=2an+n

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

[a (n+1)+(n+1)+1]/(an+n+1) = 2, which is a constant value.

a 1+ 1+ 1 = 1+ 1+ 1 = 3

The sequence {an +n+ 1} is a geometric series with 3 as the first term and 2 as the common ratio.

An+n+1= 3× 2 (n-1)

an=3×2^(n- 1) -n - 1

When n= 1, a1= 3-1-1=1also satisfies.

The general formula of the sequence {an} is an = 3× 2 (n- 1)-n- 1.

3.

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

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

1/a (n+1)-1/an = 2, which is a fixed value.

1/a 1 = 1/ 1 = 1

The sequence {1/an} is an arithmetic series, the first term is 1, and the tolerance is 2.

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

an= 1/(2n- 1)

When n= 1, a1=1(2-1) =1is also satisfied.

The general formula of sequence {an} is an= 1/(2n- 1).

4.

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

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

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

[1/a (n+1)-2]/(1/an-2) =1/2 is a fixed value.

1/a 1-2 = 1/ 1-2 = 1-2 =- 1

The sequence {1/an -2} is a geometric series with-1 as the first term and 1/2 as the common ratio.

1/ a-2 = (-1) × (1/2) (n-1) =-1/2 (n-1)

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

an=2^(n- 1)/(2？ - 1)

When n= 1, a1=1(2-1) =1is also satisfied.

The general formula of sequence {an} is an = 2 (n- 1)/(2? - 1)