设Sn为数列{an}的前n项和,Sn=(-1)nan-12n,n∈N*,则(1)a3=______;(2)S1+S2+…+S100=______.-数学

题目简介

设Sn为数列{an}的前n项和,Sn=(-1)nan-12n,n∈N*,则(1)a3=______;(2)S1+S2+…+S100=______.-数学

题目详情

设Sn为数列{an}的前n项和,Sn=(-1)nan-
1
2n
,n∈N*,则
(1)a3=______;
(2)S1+S2+…+S100=______.
题型:填空题难度:中档来源:湖南

答案

Sn=(-1)nan-class="stub"1
2n
,n∈N*,
当n=1时,有a1=(-1)1a1-class="stub"1
2
,得a1=-class="stub"1
4

当n≥2时,an=Sn-Sn-1=(-1)nan-class="stub"1
2n
-(-1)n-1an-1+class="stub"1
2n-1

an=(-1)nan+(-1)nan-1+class="stub"1
2n

若n为偶数,则an-1=-class="stub"1
2n
(n≥2)

所以an=-class="stub"1
2n+1
(n为正奇数);
若n为奇数,则an-1=-2an+class="stub"1
2n
=(-2)•(-class="stub"1
2n+1
)+class="stub"1
2n
=class="stub"1
2n-1

所以an=class="stub"1
2n
(n为正偶数).
所以(1)a3=-class="stub"1
24
=-class="stub"1
16

故答案为-class="stub"1
16

(2)因为an=-class="stub"1
2n+1
(n为正奇数),所以-a1=-(-class="stub"1
22
)=class="stub"1
22

an=class="stub"1
2n
(n为正偶数),所以a2=class="stub"1
22

-a1+a2=2×class="stub"1
22

-a3=-(-class="stub"1
24
)=class="stub"1
24
a4=class="stub"1
24

-a3+a4=2×class="stub"1
24


-a99+a100=2×class="stub"1
2100

所以,S1+S2+S3+S4+…+S99+S100
=(-a1+a2)+(-a3+a4)+…+(-a99+a100)-(class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2100
)

=2(class="stub"1
4
+class="stub"1
16
+…+class="stub"1
2100
)-(class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2100
)

=2•
class="stub"1
4
(1-class="stub"1
450
)
1-class="stub"1
4
-
class="stub"1
2
(1-class="stub"1
2100
)
1-class="stub"1
2

=class="stub"1
3
(class="stub"1
2100
-1)

故答案为class="stub"1
3
(class="stub"1
2100
-1)

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