注意等差数列的求和公式是Sn=n*(a1+an)/2=n*[a1+a1+(n-1)*d]/2=(n^2*d-n*d+2*n*a1)/2s2n/sn=(4n+2)/(n+1)=A（4n²+2n）/A（n²+n)=[d/2*（4n²）+(a1-d/2)*(2n)]/[d/2*(n^2)+(a1-d/2)*n]有d/2=a1-d/2=A=1即d=2，a1=2
