Question

两个正项数列{an}{bn},an,bn^2,a(n+1)是等差数列,bn^2,a(n+1),b(n+1)^2是等比数列

Answer 1

1.

a(n+1)^2=(bn^2)b(n+1)^2

a(n+1)=bnb(n+1)

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

=bnb(n-1)+bnb(n+1)

2bn=b(n-1)+b(n+1)

所以bn是等差数列；

2.

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

2b1^2=a1+a2=8

b1=2

a2=b1b2

b2=6/2=3

d=b2-b1=1

所以

bn=2+(n-1)=n+1

an=bnb(n-1)

=n(n+1)

=n^2-n

cn=(an-n^2)q^bn

=nq^(n+1)

Sn=q^2+2q^3+3q^4+……+(n-2)q^(n-1)+(n-1)q^n+nq^(n+1)

(1/q)Sn=q^1+2q^2+3q^3+……+(n-2)q^(n-2)+(n-1)q^(n-1)+nq^n

相减：

(1/q-1)Sn=q^1+q^2+q^3+……+q^(n-2)+q^(n-1)+q^n-nq^(n+1)

=q(1-q^n)/(1-q)-nq^(n+1)

(1/q-1)Sn=q(1-q^n)/(1-q)-nq^(n+1)

Sn=[q^2/(1-q)^2]*(1-q^n)-[1/(1-q)]nq^(n+2)

=q^2/(1-q)^2-q^(n+2)/(1-q)^2-[1/(1-q)]nq^(n+2)

=q^2/(1-q)^2-[1/(1-q)^2+n/(1-q)]q^(n+2)