摘要:(Ⅰ)解:因为{cn+1-pcn}是等比数列.故有(cn+1-pcn)2=(cn+2-pcn+1)(cn-pcn-1).将cn=2n+3n代入上式.得[2n+1+3n+1-p(2n+3n)]2=[2n+2+3n+2-p(2n+1+3n+1)]?[2n+3n-p(2n-1+3n-1)]即[(2-p)2n+(3-p)3n]2=[(2-p)2n+1+(3-p)3n+1][(2-p)2n-1+(3-p)3n-1].

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