设数列{an}首项a1=1前n项和为Sn且Sn+1=2n平方+3n+1n属N
问题描述：
设数列{an}首项a1=1前n项和为Sn且Sn+1=2n平方+3n+1n属N
求数列{an}的通项公式an
答
S(n+1)=2n^2+3n+1Sn=2(n-1)^2+3(n-1)+1=2(n^2-2n+1)+3n-3+1=2n^2-4n+2+3n-3+1=2n^2-nS(n+1)-Sn=a(n+1)=2n^2+3n+1-(2n^2-n)=2n^2+3n+1-2n^2+n=4n+1=4n+4-3=4(n+1)-3所以an=4n-3