已知数列{an},a1=p＞0,an+1乘an=n^2+3n+2,求数列{an}的通项公式

an+1乘an=n^2+3n+2=(n+2)*(n+1)则an*a(n-1)=(n+1)*n两边取对数得lgan+lga(n-1)=lg(n+1)+lgnlgan-lg(n+1)=-[lga(n-1)-lgn]所以{lgan-lg(n+1)}是公比为-1的等比数列首项=lga1-lg2=lg(p/2)所以lgan-lg(n+1)=lg(p/2)*(-1...