已知数列{an}的各项均为正数,观察程序框图,若k=5,k=10时,分别有S=511和S=1021(1)试求数列{an}的通项;(2)令bn=2an,求b1+b2+…+bm的值.-高二数学

题目简介

已知数列{an}的各项均为正数,观察程序框图,若k=5,k=10时,分别有S=511和S=1021(1)试求数列{an}的通项;(2)令bn=2an,求b1+b2+…+bm的值.-高二数学

题目详情

已知数列{an}的各项均为正数,观察程序框图,若k=5,k=10时,分别有S=
5
11
S=
10
21

(1)试求数列{an}的通项;
(2)令bn=2an,求b1+b2+…+bm的值.
题型:解答题难度:中档来源:不详

答案

(1)由框图可知
S=class="stub"1
a1a2
+class="stub"1
a2a3
+…+class="stub"1
akak+1

∵ai+1=ai+d,∴{an}是等差数列,设公差为d,则有
class="stub"1
akak+1
=class="stub"1
d
(class="stub"1
ak
-class="stub"1
ak+1
)

S=class="stub"1
d
(class="stub"1
a1
-class="stub"1
a2
+class="stub"1
a2
-class="stub"1
a3
+…+class="stub"1
ak
-class="stub"1
ak+1
)
=class="stub"1
d
(class="stub"1
a1
-class="stub"1
ak+1
)

由题意可知,k=5时,S=class="stub"5
11
;k=10时,S=class="stub"10
21

class="stub"1
d
(class="stub"1
a1
-class="stub"1
a6
)=class="stub"5
11
class="stub"1
d
(class="stub"1
a1
-class="stub"1
a11
)=class="stub"10
21
a1=1
d=2
a1=-1
d=-2
(舍去)
故an=a1+(n-1)d=2n-1
(2)由(1)可得:bn=2an=22n-1
∴b1+b2++bm=21+23++22m-1
=
2(1-4m)
1-4

=class="stub"2
3
(4m-1)

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