数列｛an｝的前n项和,且Sn=n∧²an,则an/an+1=

n≥1时,Sn=n²an=n²[Sn-S(n-1)](n²-1)Sn=n²S(n-1)Sn/S(n-1)=n²/(n²-1)=n²/[(n+1)(n-1)]S(n-1)/S(n-2)=(n-1)²/[n(n-2)]…………S2/S1=2²/(3×1)连乘Sn/S1=(2²×3²×。。。×n²)/[(3×1)(4×2)。。。(n+1)(n-1)]=(2²×3²×。。。×n²)/[3×4×。。。×(n+1)×1×2×。。。×(n-1)]=(2²×3²×。。。×n²)/[1×2×3²×4²×。。。(n-1)²×n(n+1)]=4n²/[2n(n+1)]=2n/(n+1)Sn=[2n/(n+1)]S1an=Sn-S(n-1)=[2n/(n+1) -2(n-1)/n]S1a(n+1)=S(n+1)-Sn=[2(n+1)/(n+2) -2n/(n+1)]S1an/a(n+1)=[2n/(n+1)-2(n-1)/n]/[2(n+1)/(n+2)-2n/(n+1)]=[n(n+2)n-(n-1)(n+1)(n+2)]/[(n+1)n(n+1)-n(n+2)n]=(n+2)/n=1 +2/n