Question

（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{a_n}满足：a_n+1=3f (a_n)－1(nÎ N^*)，且a₁=1，求数列{a_n}的通项公式；
(Ⅲ)求数列{a_n}的前n项和S_n．

Answer 1

(Ⅰ)  (Ⅱ) a_n = 2×3^n－1－1（Ⅲ）3ⁿ－n－2

(I) ∵f (0) =1．
令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) ∵，∴a_n+1=3f (a_n)－1= 3a_n+2,                                             
∴a_n+1+1=3(a_n+1），                                                                                 
又a₁+1=2，∴数列{a_n+1} 是公比为3的等比数列                            
∴a_n +1= 2×3^n－1，即a_n = 2×3^n－1－1                                                     10分
(III) S_n = a₁ + a₂ + … + a_n
=2×(3⁰+3¹+3²+ ×××××× + 3^n－1)－n
=3ⁿ－n－2                                                                                           14分