(理)数列{an}满足a1=1且8an+1an-16an+1+2an+5=0(n≥1)记bn=1an-12(n≥1)(1)求b1,b2,b3,b4的值.(2)求{bn}、{anbn}的通项公式.(3)

题目简介

(理)数列{an}满足a1=1且8an+1an-16an+1+2an+5=0(n≥1)记bn=1an-12(n≥1)(1)求b1,b2,b3,b4的值.(2)求{bn}、{anbn}的通项公式.(3)

题目详情

(理)数列{an}满足a1=1 且8an+1an-16an+1+2an+5=0(n≥1)记bn=
1
an-
1
2
(n≥1)

(1)求b1,b2,b3,b4的值.
(2)求{bn}、{anbn}的通项公式.
(3)求{anbn}的前n项和Sn
题型:解答题难度:中档来源:不详

答案

(1)由bn=class="stub"1
an-class="stub"1
2
an=class="stub"1
bn
+class="stub"1
2

代入8an+1an-16an+1+2an+5=0(n≥1),得8(class="stub"1
bn+1
+class="stub"1
2
)(class="stub"1
bn
+class="stub"1
2
)-16(class="stub"1
bn+1
+class="stub"1
2
)+2(class="stub"1
bn
+class="stub"1
2
)+5=0,
化简得bn+1=2bn-class="stub"4
3
,则bn+1-class="stub"4
3
=2(bn-class="stub"4
3
),
所以{bn-class="stub"4
3
}为等比数列,其公比为2,首项为b1-class="stub"4
3
=class="stub"1
a1-class="stub"1
2
-class="stub"4
3
=class="stub"2
3

所以bn-class="stub"4
3
=class="stub"2
3
•2n-1=
2n
3

所以bn=
2n
3
+class="stub"4
3

所以b1=class="stub"2
3
+class="stub"4
3
=2,b2=
22
3
+class="stub"4
3
=class="stub"8
3
b3=
23
3
+class="stub"4
3
=4,b4=
24
3
+class="stub"4
3
=class="stub"20
3

(2)由(1)求解过程可知bn=
2n
3
+class="stub"4
3

an=class="stub"1
bn
+class="stub"1
2
=class="stub"3
2n+4
+class="stub"1
2

所以anbn=(class="stub"3
2n+4
+class="stub"1
2
)(
2n
3
+class="stub"4
3
)=1+
2n-1+2
3
=class="stub"5
3
+
2n-1
3

(3)Sn=(class="stub"5
3
+class="stub"1
3
)+(class="stub"5
3
+class="stub"2
3
)+…+(class="stub"5
3
+
2n-1
3
)=class="stub"5
3
n+
class="stub"1
3
(1-2n)
1-2
=class="stub"5
3
n+
2n-1
3

更多内容推荐