a1= p+q+1,an = 2n+p-1 ,n≥2 ；1) 由于{an}是等差数列,必有 p+q+1 = 2×1 +p-1,解得：q=0；所以 a2= p+3,a3=p+5,a5=p+9 ,由等比数列解得：p= -1；2) an = 2n-2,则 bn = n × 2^(2n-2) = n × 4^(n-1) ,由错位相减法...