an的前n项和Sn,a1=7,an+1=(n+2)/nSn,证数列Sn/n是等比数列和S(n+1)=4an

令Sn/n=bn则a(n+1)=Sn+2*bn,(n+1)*b(n+1)-n*bn=n*bn+2*bn,b(n+1)=2*bn故bn是等比数列第二问,相当于要求证明S(n+2)=4*S(n+1)-4Sn即(n+2)*b(n+2)=4*(n+1)*b(n+1)-4*n*bn=2*(n+1)*b(n+2)-n*b(n+2)显然成立,证明时反推即...