已知数列{an}中,Sn是它的前n项和,并且Sn+1=4an＋2(n∈N*),a1=1

(1)S n+1 =4a n ＋2 Sn=4an-1+2 两式相减 an+1=4(an-an-1) an+1-2an=2(an-an-1) b n =a n+1 －2a n bn=2bn-1 数列{b n }是等比数列 (2)a2=5 b1=5-2=3 bn=3*(2)^(n-1) an+1-2an=3*(2)^(n-1) 两边各除以2^(n+1) a(n+1)/2^(n+1)-an/2^n=3/4 a(1)/2=1/2 an/2^n=(3n-1)/4 an=2^n*(3n-1)/4