已知数列{an}满足an+1=2an+3.5^n,a1=6.求an

a(n+1)-2an=3。5^n,则a2-2a1=3。5^1a3-2a2=3。5^2。a(n+1)-2an=3。5^n以上式子相加,得a(n+1)-a1-Sn=3。5+3。5^2+。。。+3。5^n=3。5(1-3。5^n)/(1-3。5)=7(3。5^n-1)/5a(n+1)=S(n+1)-Sn--->a(n+1)-S(n+1)=-Sn-Sn-a1=7(3。5^n-1)/5--->Sn=-a1-7(3。5^n-1)/5=-6-7(3。5^n-1)/5an=Sn-S(n-1)=[-6-7(3。5^n-1)/5]-[-6-7(3。5^(n-1)-1)/5]=[7*3。5^(n-1)-7*3。5^n]/5=-3。5^n有点怪异!