问题补充:
设数列满足a1=2,an+1-an=3?22n-1
(1)求数列{an}的通项公式;
(2)令bn=nan,求数列的前n项和Sn.
答案:
解:(Ⅰ)由已知,当n≥1时,an+1=[(an+1-an)+(an-an-1)++(a2-a1)]+a1
=3(22n-1+22n-3+…+2)+2=22(n+1)-1.
而a1=2,
所以数列{an}的通项公式为an=22n-1.
(Ⅱ)由bn=nan=n?22n-1知Sn=1?2+2?23+3?25+…+n?22n-1①
从而22Sn=1?23+2?25+…+n?22n+1②
①-②得(1-22)?Sn=2+23+25+…+22n-1-n?22n+1.
即.
解析分析:(Ⅰ)由题意得an+1=[(an+1-an)+(an-an-1)+…+(a2-a1)]+a1=3(22n-1+22n-3+…+2)+2=22(n+1)-1.由此可知数列{an}的通项公式为an=22n-1.(Ⅱ)由bn=nan=n?22n-1知Sn=1?2+2?23+3?25++n?22n-1,,由此入手可知