题目内容
对于数列{an},定义数列{an+1-an}为数列{an}的“差数列”,若a1=2,{an}的“差数列”的通项为2n,则数列{an}的前n项和Sn=______.
∵an+1-an=2n,
∴an=(an-an-1)+(an-1-an-2)++(a2-a1)+a1
=2n-1+2n-2++22+2+2
=
+2=2n-2+2=2n.
∴Sn=
=2n+1-2.
故答案为2n+1-2
∴an=(an-an-1)+(an-1-an-2)++(a2-a1)+a1
=2n-1+2n-2++22+2+2
=
| 2-2n |
| 1-2 |
∴Sn=
| 2-2n |
| 1-2 |
故答案为2n+1-2
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