设数列{An}中,A1=2,An+1=An+2n+1,则通项公式An=

a(n+1)=an+2n+1a(n+1)-an=+2n+1an-a(n-1)=2(n-1)+1.a3-a2=2*2+1a2-a1=2*1+1以上等式相加得a(n+1)-a1=2*1+1+2*2+1+.+2n+1a(n+1)-a1=2*(1+2+3+...+n)+na(n+1)-a1=n(n+1)+na(n+1)-a1=n^2+2na(n+1)=n^2+2n-2a(n+1)=(n+1)...