已知数列an,a1=2,且a（n+1）=2an+3n,求an

因为a（n+1）=2an+3n令a(n+1)+x(n+1)+y=2[an+xn+y]则a(n+1)=2an+xn+y-x所以x=3,y-x=0故x=3,y=3所以a(n+1)+3(n+1)+3=2[an+3n+3]所以数列{an+3n+3}是等比数列,首项是a1+3+3=8,公比是q=2所以an+3n+3=8*2^(n-1)=2^(n+2)...