已知等差数列{an}是递增数列,且满足a4•a7=15,a3+a8=8.(1)求数列{an}的通项公式;(2)令bn=19an-1an(n≥2),b1=13,求数列{bn}的前n项和Sn.-数学

题目简介

已知等差数列{an}是递增数列,且满足a4•a7=15,a3+a8=8.(1)求数列{an}的通项公式;(2)令bn=19an-1an(n≥2),b1=13,求数列{bn}的前n项和Sn.-数学

题目详情

已知等差数列{an}是递增数列,且满足a4•a7=15,a3+a8=8.
(1)求数列{an}的通项公式;
(2)令bn=
1
9an-1an
(n≥2),b1=
1
3
,求数列{bn}的前n项和Sn
题型:解答题难度:中档来源:不详

答案

(1)根据题意:a3+a8=8=a4+a7,a4•a7=15,知:a4,a7是方程x2-8x+15=0的两根,且a4<a7
解得a4=3,a7=5,设数列{an}的公差为d
a7=a4+(7-4)•d,得d=class="stub"2
3
.

故等差数列{an}的通项公式为:an=a4+(n-4)•d=3+(n-4)•class="stub"2
3
=class="stub"2n+1
3

(2)bn=class="stub"1
9an-1an
=class="stub"1
9(class="stub"2
3
n-class="stub"1
3
)(class="stub"2
3
n+class="stub"1
3
)
=class="stub"1
(2n-1)(2n+1)
=class="stub"1
2
(class="stub"1
2n-1
-class="stub"1
2n+1
)

b1=class="stub"1
3
=class="stub"1
2
(1-class="stub"1
3
)

Sn=b1+b2++bn=class="stub"1
2
(1-class="stub"1
3
+class="stub"1
3
-class="stub"1
5
++class="stub"1
2n-1
-class="stub"1
2n+1
)=class="stub"1
2
(1-class="stub"1
2n+1
)
=class="stub"n
2n+1

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