问题 解答题
已知数列{an}满足a1=1,an+1=
1
2
an+n,n为奇数
an-2n,n为偶数
,记bn=a2n,n∈N*
(1)求a2,a3
(2)求数列{bn}的通项公式;
(3)求S2n+1
答案

(1)当n=2时,a2=

1
2
a1+1=
1
2
+1=
3
2

当n=3时,a3=a2-2×2=

3
2
-4=-
5
2

(2)当n≥2时,bn=a2n=a(2n-1)+1=

1
2
a2n-1+(2n-1)

=

1
2
[a2n-2-2(2n-2)]+(2n-1)=
1
2
a2(n-1)+1=
1
2
bn-1+1

∴bn-2=

1
2
(bn-1-2),又b1-2=a2-2=-
1
2

∴bn-2=-

1
2
•(
1
2
n-1=-(
1
2
n,即bn=2-(
1
2
n

(3)∵a2n+1=a2n-4n=bn-4n

∴S2n+1=a1+a2+…+a2n+a2n+1

=(a2+a4+…+a2n)+(a1+a3+a5+…+a2n+1

=(b1+b2+…+bn)+[a1+(b1-4×1)+(b2-4×2)+…+(bn-4×n)]

=a1+2(b1+b2+…+bn)-4×(1+2+…+n)

=1+2(2n-

1
2
[1-(-
1
2
)
n
]
1-
1
2
)-4×
n(n+1)
2

=(

1
2
n-1-2n2+2n-1.

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