Question

数列{an}的前n项和为Sn,已知a1=1/2,Sn=n^2-n(n-1)(n=1,2,3.)写出Sn与Sn+1的递推关

Answer 1

1.Sn=n*an-n(n-1)

Sn-1=(n-1)an-1-(n-2)(n-1) n>1

前式减后式

an=n*an-(n-1)an-1-2(n-1)

(n-1)*an-(n-1)an-1-2(n-1)=0

(n-1)(an-an-1-2)=0 n>1

an-an-1=2 n>1

数列(an)是公差为2的等差数列

an=1/2+2(n-1)=2n-3/2

S1=a1=1/2

S2=1/2+1/2+2=3

S3=1/2+1/2+2+1/2+4=15/2

Sn=(1/2+2n-3/2)n/2

=(2n-1)n/2

2.Fn(x)=(Sn/n)x^(n+1)=nx^(n+1)/(n+1)

F'n(p)=n(n+1)p^n/(n+1)=np^n=Bn

若p=1,则Bn=n,则Tn=n(n+1)/2；

若p≠1,这是个很熟悉的关系式,利用错位相减：

Tn=p+2p²+……+np^n

pTn=p²+2p^3+……+np^(n+1)

两式相减=(p-1)Tn=np^(n+1)-(p+p²+……+p^n)=np^(n+1)-p(1-p^n)/(1-p)

Tn=[np^(n+1)-p(1-p^n)/(1-p)]/(p-1)