a、b∈R,且|a|<1,|b|<1,则无穷数列:1,(1+b)a,(1+b+b2)a2,…,(1+b+b2+…+bn-1)an-1…的和
a、b∈R,且|a|<1,|b|<1,则无穷数列:1,(1+b)a,(1+b+b2)a2,…,(1+b+b2+…+bn-1)an-1…的和为()A.1(1?a)(1?b)...
a、b∈R,且|a|<1,|b|<1,则无穷数列:1,(1+b)a,(1+b+b2)a2,…,(1+b+b2+…+bn-1)an-1…的和为( )A.1(1?a)(1?b)B.11?abC.2(1?a)(1?ab)D.1(1?a)(1?ab)
展开
1个回答
展开全部
∵an=(1+b+b2+…+bn-1)an-1
=
?an-1
=
(an-1-an-1bn),
∵|a|<1,|b|<1,
∴无穷数列:1,(1+b)a,(1+b+b2)a2,…,(1+b+b2+…+bn-1)an-1…的和:
S=
[
-
]
=
[
-
-
+
?(ab)n]
=
-
=
=
,
故选D.
=
1?bn |
1?b |
=
1 |
1?b |
∵|a|<1,|b|<1,
∴无穷数列:1,(1+b)a,(1+b+b2)a2,…,(1+b+b2+…+bn-1)an-1…的和:
S=
lim |
n→∞ |
1 |
1?b |
1?an |
1?a |
b(1?anbn) |
1?ab |
=
lim |
n→∞ |
1 |
(1?b)(1?a) |
an |
(1?b)(1?a) |
b |
(1?b)(1?ab) |
b |
(1?a)(1?ab) |
=
1 |
(1?b)(1?a) |
b |
(1?b)(1?ab) |
=
1?ab?b(1?a) |
(1?b)(1?a)(1?ab) |
=
1 |
(1?a)(1?ab) |
故选D.
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询