数列{an}的前n项和为Sn=npan(n∈N*)且a1≠a2,(1)求常数p的值;(2)证明:数列{an}是等差数列.-数学

题目简介

数列{an}的前n项和为Sn=npan(n∈N*)且a1≠a2,(1)求常数p的值;(2)证明:数列{an}是等差数列.-数学

题目详情

数列{an}的前n项和为Sn=npan(n∈N*)且a1≠a2
(1)求常数p的值;
(2)证明:数列{an}是等差数列.
题型:解答题难度:中档来源:不详

答案

(1)当n=1时,a1=pa1,若p=1时,a1+a2=2pa2=2a2,
∴a1=a2,与已知矛盾,故p≠1.则a1=0.
当n=2时,a1+a2=2pa2,∴(2p-1)a2=0.
∵a1≠a2,故p=class="stub"1
2

(2)由已知Sn=class="stub"1
2
nan,a1=0.
n≥2时,an=Sn-Sn-1=class="stub"1
2
nan-class="stub"1
2
(n-1)an-1.
an
an-1
=class="stub"n-1
n-2
.则
an-1
an-2
=class="stub"n-2
n-3
a3
a2
=class="stub"2
1

an
a2
=n-1.∴an=(n-1)a2,an-an-1=a2.
故{an}是以a2为公差,以a1为首项的等差数列.

更多内容推荐