18.已知函数f(x)=.数列{an}满足a1=1.an+1=f(an)(n∈N*). (1)求证:数列{}是等差数列, (2)记Sn(x)=++-+eq \f(xn,an).求Sn(x). (1)证明:∵an+1=f(an).∴an+1=. ∴=+3.即-=3. ∴{}是以=1为首项.3为公差的等差数列. ∴=1+3(n-1)=3n-2. (2)解:Sn(x)=x+4x2+7x3+-+(3n-2)xn.① 当x=1时.Sn(x)=1+4+7+-+(3n-2)==. 当x≠1时.xSn(x)=x2+4x3+-+(3n-5)xn+(3n-2)xn+1.② ①-②.得(1-x)Sn(x)=x+3x2+3x3+-+3xn-(3n-2)xn+1=3(x+x2+-+xn)-2x-(3n-2)xn+1=-2x-(3n-2)xn+1. Sn(x)=-. 查看更多

 

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已知函数f(x)=,数列{an}满足a1=1,an+1=f(an),n=1,2,3……

(1)求证:数列为等比数列,并求数列{an}的通项公式an

(2)若数列{bn}满足bnanan+1·3n,Sn为数列{bn}的前n项和,求Sn

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已知函数f(x)=,数列{an}满足a1=1,an+1=f(an)(n∈N*)

(1)求数列{an}的通项公式;

(2)记Sn=a1a2+a2a3+……+anan+1,求Sn

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已知函数f(x)=,数列{an}满足a1=1,an+1=f(an)(n∈N*).

(1)求数列{an}的通项公式;

(2)记Sn=a1a2+a2a3+…+an+an+1,求Sn

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已知函数f(x)=,数列{an}满足:a1,an+1=f(an).

(1)求证数列为等差数列,并求数列{an}的通项公式;

(2)记Sn=a1a2+a2a3+…+anan+1,求证:Sn

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已知函数f(x)=,数列{an}满足:a1,an+1=f(an)

(1)求证:数列{}为等差数列,并求数列{an}的通项公式;

(2)记Sn=a1a2+a2a3+……+anan+1,求证:Sn

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