Question

等差数列an中,a1=1前n项和Sn,满足条件S2n/Sn=4n+2/n+1,求an通项 记Bn=anp^an(p>0)

Answer 1

1)因为Sn=na1+n(n-1)d/2=n+n(n-1)d/2,S2n=2n+2n(2n-1)d/2,

S（2n）/Sn=（4n+2）/（n+1）,所以d=1,所以Sn=n+n(n-1)/2

2)an=n,所以bn=n*p^n,

bn=p*b(n-1)+p^n

b(n-1)=p*b(n-2)+p^(n-1)

b(n-2)=p*b(n-3)+p^(n-2)

.

b2=p*b1+p^2

相加起来得到：Tn-b1=p*[T(n-1)]+p^2[1-p^(n-1)]/(1-p)

Tn-b1=p*[Tn-bn]+p^2[1-p^(n-1)]/(1-p)

Tn(1-p)=b1-bn*p+p^2[1-p^(n-1)]/(1-p)

Tn={p-[n*p^n]*p+p^2[1-p^(n-1)]/(1-p)}/(1-p)

化简得到：Tn=[1-p^(n+1)+n*(p-1)*p^(n+1)]/(1-p)^2