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,证明：（1）{bn}是等差数列（2）若a1=2,a2=6,设cn=(an-n^2)q^bn(q>0,为常数),求{cn}前n项和Sn

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,证明：（1）{bn}是等差数列（2）若a1=2,a2=6,设cn=(an-n^2)q^bn(q>0,为常数),求{cn}前n项和Sn
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,证明：
（1）{bn}是等差数列
（2）若a1=2,a2=6,设cn=(an-n^2)q^bn(q>0,为常数),求{cn}前n项和Sn

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,证明：（1）{bn}是等差数列（2）若a1=2,a2=6,设cn=(an-n^2)q^bn(q>0,为常数),求{cn}前n项和Sn
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)

an,bn^2,a(n+1)是等差数列 => 2bn^2 = an + a(n+1)
bn^2,a(n+1),b(n+1)^2是等比数列 => a(n+1) ^ 2 = bn^2 * b(n+1)^2 => a(n+1) = bn * b(n+1)
所以，2bn^2 = b(n-1) * bn + bn* b(n+1) => 2bn = b(n-1) + b(n+1)
第二...

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an,bn^2,a(n+1)是等差数列 => 2bn^2 = an + a(n+1)
bn^2,a(n+1),b(n+1)^2是等比数列 => a(n+1) ^ 2 = bn^2 * b(n+1)^2 => a(n+1) = bn * b(n+1)
所以，2bn^2 = b(n-1) * bn + bn* b(n+1) => 2bn = b(n-1) + b(n+1)
第二问an bn通项都有了那么cn也就很简单了吧，cn的通项应该可以用q * cn - cn这种方法解

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