首页 > 已知a=limn→+∞(1n2+2n2+…+nn2)，b=limn→+∞(1+13+19+…+13n-1+…)，则a、b的值分别为______，c=limn→+∞an+bnan+1+bn+1=____

已知a=limn→+∞(1n2+2n2+…+nn2)，b=limn→+∞(1+13+19+…+13n-1+…)，则a、b的值分别为______，c=limn→+∞an+bnan+1+bn+1=____

题目简介

已知a=limn→+∞(1n2+2n2+…+nn2)，b=limn→+∞(1+13+19+…+13n-1+…)，则a、b的值分别为______，c=limn→+∞an+bnan+1+bn+1=____

题目详情

已知a=
lim
n→+∞
(
1
n2
+
2
n2
+…+
n
n2
)，b=
lim
n→+∞
(1+
1
3
+
1
9
+…+
1
3n-1
+…)，则a、b的值分别为______，c=
lim
n→+∞
an+bn
an+1+bn+1
=______．
题型：填空题难度：偏易来源：朝阳区一模

答案

∵class="stub"1
n2
+class="stub"2
n2
+…+class="stub"n
n2
=
n(n+1)
2
n2
=class="stub"n+1
2n
，∴a=
lim
n→∞
class="stub"n+1
2n
=
lim
n→∞
1+class="stub"1
n
2
=class="stub"1
2
；
∵1+class="stub"1
3
+class="stub"1
9
+…+class="stub"1
3n-1
=
1-class="stub"1
3n
1-class="stub"1
3
，∴b=
lim
n→∞
1-class="stub"1
3n
1-class="stub"1
3
=class="stub"3
2
；
则
an+bn
an+1+bn+1
=
class="stub"1
2n
+(class="stub"3
2
)n
class="stub"1
2n+1
+(class="stub"3
2
)n+1
=
class="stub"1
3n
+1
class="stub"1
2
×class="stub"1
3n
+class="stub"3
2
，
所以c=
lim
n→∞
class="stub"1
3n
+1
class="stub"1
2
×class="stub"1
3n
+class="stub"3
2
=class="stub"2
3
，
故答案为：class="stub"1
2
，class="stub"3
2
；class="stub"2
3
．

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