问题
解答题
已知数列{an}满足a1=1,an+1=
(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=
a1+1=1 2
+1=1 2
;3 2
当n=3时,a3=a2-2×2=
-4=-3 2
.5 2
(2)当n≥2时,bn=a2n=a(2n-1)+1=
a2n-1+(2n-1)1 2
=
[a2n-2-2(2n-2)]+(2n-1)=1 2
a2(n-1)+1=1 2
bn-1+11 2
∴bn-2=
(bn-1-2),又b1-2=a2-2=-1 2
,1 2
∴bn-2=-
•(1 2
)n-1=-(1 2
)n,即bn=2-(1 2
)n.1 2
(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-
)-4×
[1-(-1 2
)n]1 2 1- 1 2 n(n+1) 2
=(
)n-1-2n2+2n-1.1 2