数列{an}的前n项和为Sn,满足Sn=32an-n2-34,设bn=log3(an+12),则数列{1bn•bn+1}的前19项和为______.-数学

题目简介

数列{an}的前n项和为Sn,满足Sn=32an-n2-34,设bn=log3(an+12),则数列{1bn•bn+1}的前19项和为______.-数学

题目详情

数列{an}的前n项和为Sn,满足Sn=
3
2
an-
n
2
-
3
4
,设bn=log3(an+
1
2
)
,则数列{
1
bnbn+1
}
的前19项和为 ______.
题型:填空题难度:偏易来源:大连模拟

答案

令n=1,得到a1=s1=class="stub"3
2
a1-class="stub"1
2
-class="stub"3
4
,解得a1=class="stub"5
2

因为sn=class="stub"3
2
an-class="stub"n
2
-class="stub"3
4

当n≥2时求出sn-1=class="stub"3
2
an-1-class="stub"n-1
2
-class="stub"3
4

用①-②得:an=3an-1+1,所以代入求得a2=class="stub"17
2
,a3=class="stub"53
2
,a4=class="stub"161
2
,…
所以数列{an+class="stub"1
2
}为以3为首项,3为公比的等比数列,
所以通项公式为3n,则bn=
log3n3
=n,
数列{class="stub"1
bnbn+1
}
的前19项和为:
class="stub"1
b1b2
+class="stub"1
b2b3
+…+class="stub"1
b19b20
=class="stub"1
1×2
+class="stub"1
2×3
+..+class="stub"1
19×20

=1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
18
-class="stub"1
19
+class="stub"1
19
-class="stub"1
20

=class="stub"19
20

故答案为class="stub"19
20

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