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)