已知数列{a n }满足:a 1 =1,a n+1 = 1 2 a n +n(n为奇数) a
已知数列{an}满足:a1=1,an+1=12an+n(n为奇数)an-2n(n为偶数)(1)求a2,a3,a4,a5;(2)设bn=a2n+1+4n-2,n∈N*,求证...
已知数列{a n }满足:a 1 =1,a n+1 = 1 2 a n +n(n为奇数) a n -2n(n为偶数) (1)求a 2 ,a 3 ,a 4 ,a 5 ; (2)设b n =a 2n+1 +4n-2,n∈N * ,求证:数列{b n }是等比数列,并求其通项公式; (3) 求数列{a n }前100项中的所有奇数项的和S.
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(1)
a
2
=
3
2
,
a
3
=-
5
2
,
a
4
=
7
4
,
a
5
=-
25
4
(2)b
n+1
=a
2n+3
+4(n+1)-2=a
2n+2
-2(2n+2)+4(n+1)-2
=
a
2n+2
-2=
1
2
a
2n+1
+(2n+1)-2=
1
2
b
n
∴数列{b
n
}是公比为
1
2
的等比数列.
又∵
b
1
=
a
3
+4-2=-
1
2
,∴
b
n
=-
(
1
2
)
n
(3)由(2)得
a
2n+1
=-
(
1
2
)
n
-4n+2
∴s=a
1
+a
3
+…+a
99
=1-[
1
2
+
(
1
2
)
2
+
(
1
2
)
3
+…+
(
1
2
)
49
]-4(1+2+…+49)+2×49
=
(
1
2
)
49
-4802
a
2
=
3
2
,
a
3
=-
5
2
,
a
4
=
7
4
,
a
5
=-
25
4
(2)b
n+1
=a
2n+3
+4(n+1)-2=a
2n+2
-2(2n+2)+4(n+1)-2
=
a
2n+2
-2=
1
2
a
2n+1
+(2n+1)-2=
1
2
b
n
∴数列{b
n
}是公比为
1
2
的等比数列.
又∵
b
1
=
a
3
+4-2=-
1
2
,∴
b
n
=-
(
1
2
)
n
(3)由(2)得
a
2n+1
=-
(
1
2
)
n
-4n+2
∴s=a
1
+a
3
+…+a
99
=1-[
1
2
+
(
1
2
)
2
+
(
1
2
)
3
+…+
(
1
2
)
49
]-4(1+2+…+49)+2×49
=
(
1
2
)
49
-4802
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