已知数列{an}的前n项和的公式是Sn=π12(2n2+n).(1)求证:{an}是等差数列,并求出它的首项和公差;(2)记bn=sinan•sinan+1•sinan+2,求出数列{an•bn}的前

题目简介

已知数列{an}的前n项和的公式是Sn=π12(2n2+n).(1)求证:{an}是等差数列,并求出它的首项和公差;(2)记bn=sinan•sinan+1•sinan+2,求出数列{an•bn}的前

题目详情

已知数列{an}的前n项和的公式是Sn=
π
12
(2n2+n)

(1)求证:{an}是等差数列,并求出它的首项和公差;
(2)记bn=sinan•sinan+1•sinan+2,求出数列{an•bn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

当n=1时,
a1=S1=class="stub"π
4

当n≥2时,an=Sn-Sn-1=class="stub"π
12
(2n2+n)-class="stub"π
12
[2(n-1)2+(n-1)]
=class="stub"π
12
(4n-1)

所以an=class="stub"π
12
(4n-1)
.an-a n-1=class="stub"π
3
,所以{an}是等差数列,它的首项为class="stub"π
4
和公差为class="stub"π
3

(2)b1=sina1•sina2•sina3=sinclass="stub"π
4
sinclass="stub"7π
12
sinclass="stub"11π
12
=
2
2
×(-class="stub"1
2
)×(cosclass="stub"18π
12
-cosclass="stub"4π
12
)=
2
8

bn
bn-1
=
sinan-2
sinan-1
=
sin(an-1+π)
sinan-1
=
-sinan-1
sinan-1
=-1,数列{bn}是等比数列,首项为
2
8
,公比为-1.
所以bn=
2
8
(-1)n-1
,anbn=
2
π
96
(-1)n-1(4n-1)

错位相减法得Tn=
2
π
192
[1-(-1)n(4n+1)]

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