Question

设数列{求an}的前n项和为Sn，已知a1=a，Sn+1=2Sn+n+1 数列{an}的同乡公式

Answer 1

解：1）(Sn+1)
+
n
+
3
=
2[(Sn)
+
n
+
2]，而且S1
+
3
=
a
+
3
当a
≠
-3时，数列{(Sn)
+
n
+
2}构成一个以a
+
3为首项，2为公比的等比数列，因此(Sn)
+
n
+
2
=
(a
+
3)*2
n
–
1
=>
Sn
=
(a
+
3)*2
n
–
1
–
n
–
2
=>
当n
≥
2，n∈N时，an
=
Sn
–
Sn-1
=
(a
+
3)
*2
n
–
2
–
1
检查a1是否符合上式，可得，
当
a
=
1
时，
an
=
(a
+
3)
*2
n
–
2
–
1
=
2
n
–
1
，
(n
∈
Z
+
)
；
当
a
≠
1
而且
a
≠
-3
时，
an
={
a
，
(n
=
1)

{
(a
+
3)
*2
n
–
2
–
1
，
(n
≥
2
，
n
∈
N)
；
当a
=
-
3时，S
2
+
4
=
0
=>
S
2
=
-4
=>
a2
=
-1
；S
3
=
2S
2
+
3
=
-5
=>
a3
=
-1，Sn+1
=
(2Sn)
+
n
+
1
=>
当n
≥
2，n∈N时，Sn
=
(2Sn-1)
+
n
=>
an+1
=
2an
+
1
=>
(an+1)
+
1
=
2[(an)
+
1]，而a2
=
-1，所以以后的每一项都是
-1
，可得
当
a
=
-
3
时，
an
=
{
-3
，
(n
=
1)

{
-1
，
(n
≥
2
，
n
∈
N)
；
2）当a
=
1时，an
=
2
n
–
1
=>
bn
=
n/an+1-an
=
n/[2
n+1
–
2
n
]
=
n/2
n
所以{bn}的前n项和为1/2
1
+
2/2
2
+
3/2
3
+
……
+
(n
-
1)/2
n
-1
+
n/2
n
=
(1/2
1
+
1/2
2
+
1/2
3
+
……
+
1/2
n
-1
+
1/2
n
)
+
[1/2
2
+
2/2
3
+
……
+
(n
-
2)/2
n
-1
+
(n
-
1)/2
n
]
=
(1/2)*[1
-
(1/2)
n
]
/
[1
–
(1/2)]
+
[1/2
2
+
2/2
3
+
……
+
(n
-
2)/2
n
-1
+
(n
-
1)/2
n
]
=
1
–
(1/2)
n
+
[1/2
2
+
2/2
3
+
……
+
(n
-
2)/2
n
-1
+
(n
-
1)/2
n
]
=
1
–
(1/2)
n
+
[1/2
2
+
2/2
3
+
……
+
(n
-
2)/2
n
-1
+
(n
-
1)/2
n
]
=
……
=
1
–
(1/2)
n
+
(1/2)[1
–
(1/2)
n
-
1
]
+
(1/4)
[1
–
(1/2)
n
-
2
]
+
……
+
(1/2)
n-2
[1
–
(1/2)
2
]
+
(1/2)
n-1
[1
-
(1/2)
1
]
=
1
+
(1/2)
+
(1/2)
2
+
……
+
(1/2)
n-1
–
n(1/2)
n
=
1[1
–
(1/2)
n
]/
[1
–
(1/2)]
–
n(1/2)
n
=
2[1
–
(1/2)
n
]
–
n(1/2)
n
=
2
–
(n
+2)
(1/2)
n
<
2
，得证。