数列{an}中,已知a1=1,an+1=an+n+12n,求an.-数学

题目简介

数列{an}中,已知a1=1,an+1=an+n+12n,求an.-数学

题目详情

数列{an}中,已知 a1=1,an+1=an+
n+1
2n
,求an
题型:解答题难度:中档来源:不详

答案

由已知可得:an+1-an=class="stub"n+1
2n

即   a2-a1=class="stub"2
21
  
  a3-a2=class="stub"3
22


an-an-1=class="stub"n
2n-1
  (n≥2)
 叠加后可得:an-a1=class="stub"2
21
+class="stub"3
22
+class="stub"4
23
+…+class="stub"n
2n-1

设 S=class="stub"2
21
+class="stub"3
22
+class="stub"4
23
+…+class="stub"n
2n-1
  (1)
则  2S=class="stub"2
20
+class="stub"3
21
+class="stub"4
22
+…+class="stub"n
2n-2
 (2)
(2)-(1)得:S=2+class="stub"1
21
+class="stub"1
22
+class="stub"1
23
+…+class="stub"1
2n-2
-class="stub"n
2n-1

=2+
class="stub"1
2
(1-class="stub"1
2n-2
)
1-class="stub"1
2
-class="stub"n
2n-1
=3-class="stub"n+2
2n-1

则 an=4-class="stub"n+2
2n-1
 (n≥2)对n=1时也符合.
故an=4-class="stub"n+2
2n-1
(n≥1)

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