(1)设bn=cn+1-pcn=2^(n+1)+3^(n+1)-p(2^n+3^n)=(2-p)*2^n+(3-p)*3^nb1=2(2-p)+3(3-p)=13-5pb2=4(2-p)+9(3-p)=35-13pb3=8(2-p)+27(3-p)=97-35p因为bn为G.P∴(35-13p)?=(13-5p)(97-35p)解得p1=2,p2=3∴p=2或p=3(2)设an=ax^(n-1),bn=by^(n-1)∴cn=an+bn=ax^(n-1)+by^(y-1)c1=a+bc2=ax+byc3=ax^2+by^2假设cn为等比,则c2^2=c1*c3∴(ax+by)?=(a+b)(ax?+by?)化简得2abxy=ab(x?+y?)∴x?+y?=2xy,(x-y)?=0,x=y与已知条件x≠y矛盾,所以假设不成立
所以cn不是等比数列