(本小题满分14分)设函数f(x)满足f(0)=1,且对任意,都有f(xy+1)=f(x)f(y)-f(y)-x+2.(I)求f(x)的解析式;(II)若数列{an}满足:an+1=3f(an)-1(

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(本小题满分14分)设函数f(x)满足f(0)=1,且对任意,都有f(xy+1)=f(x)f(y)-f(y)-x+2.(I)求f(x)的解析式;(II)若数列{an}满足:an+1=3f(an)-1(

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(本小题满分14分)设函数f (x)满足f (0) =1,且对任意,都有f (xy+1) = f (x) f (y)-f (y)-x+2.(I)      求f (x) 的解析式;(II)  若数列{an}满足:an+1=3f (an)-1(nÎ N*),且a1=1,求数列{an}的通项公式;
(Ⅲ)求数列{an}的前n项和Sn
题型:解答题难度:偏易来源:不详

答案

(Ⅰ)  (Ⅱ) an = 2×3n-1-1(Ⅲ)3nn-2
(I) ∵f (0) =1.
x=y=0得f (1) = f (0) f (0)-f (0)-0+2="2                                       "
再令y=0得,                              
所以                                                                          5分
(II) ∵,∴an+1=3f (an)-1= 3an+2,                                             
an+1+1=3(an+1),                                                                                 
a1+1=2,∴数列{an+1} 是公比为3的等比数列                            
an +1= 2×3n-1,即an = 2×3n-1-1                                                     10分
(III) Sn = a1 + a2 + … + an
=2×(30+31+32+ ×××××× + 3n-1)-n
=3nn-2                                                                                           14分

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