已知数列{an}中a1=1,a2=2,数列{an}的前n项和为Sn,当整数n>1时,Sn+1+Sn-1=2(Sn+S1)都成立,则数列{1anan+1}的前n项和为______.-数学

题目简介

已知数列{an}中a1=1,a2=2,数列{an}的前n项和为Sn,当整数n>1时,Sn+1+Sn-1=2(Sn+S1)都成立,则数列{1anan+1}的前n项和为______.-数学

题目详情

已知数列{an}中a1=1,a2=2,数列{an}的前n项和为Sn,当整数n>1时,Sn+1+Sn-1=2(Sn+S1)都成立,则数列{
1
anan+1
}的前n项和为______.
题型:填空题难度:偏易来源:不详

答案

由于a1=1,a2=2,当整数n>1时,Sn+1+Sn-1=2(Sn+S1)都成立,
所以S1=a1=1,S2=3,S3=7,故a3=4,
由于数列{an}中数列{an}的前n项和为Sn,当整数n>1时,Sn+1+Sn-1=2(Sn+S1)都成立,
则Sn+2+Sn=2(Sn+1+S1)所以an+2+an=2an+1,则数列{an}从第二项起为等差数列,
则数列an=
1,n=1
2n-2,n≥2
,所以n>1时,class="stub"1
anan+1
=class="stub"1
(2n-2)(2(n+1)-2)
=class="stub"1
2n-2
-class="stub"1
2(n+1)-2
=class="stub"1
2n-2
-class="stub"1
2n

故数列{class="stub"1
anan+1
}的前n项和为Tn=(1-class="stub"1
2
)+class="stub"1
2
[(class="stub"1
2
-class="stub"1
4
)+(class="stub"1
4
-class="stub"1
6
)…+(class="stub"1
2n-2
-class="stub"1
2n
)]
=class="stub"1
2
+class="stub"1
2
(class="stub"1
2
-class="stub"1
2n
)
=class="stub"3n-1
4n

故答案为class="stub"3n-1
4n

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