设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※).
问题描述:
设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※).
(1)令bn=an+1-an,(n属于N※),求数列(bn)的通项公式.(2)求数列{nan}的前n项和Sn
答
(I)因bn=an+1-an,所以bn+1=an+2-an+1=5/3an+1-2/3an-an+1=2/3(an+1-an)=2/3bn故{bn}是公比为2/3的等比数列,且b1=a2-a1=2/3 \x09(II)由 bn=an+1-an得an+1-a1=(an+1-an)+(an-an-1)+……+(a2-a...