己知数列{an}满足:a1=1,an+1=
(1)求a2,a3; (2)设bn=a2n-2,n∈N*,求证{bn} 是等比数列,并求其通项公式; (3)在(2)条件下,求数列{an} 前100项中的所有偶数项的和S. |
(Ⅰ)由题意可得,a2=
a1+1=1 2
×1+1=1 2
,a3=a2-4=-3 2
,(4分)5 2
(Ⅱ)∵
=bn+1 bn
=a2n+2-2 a2n-2
a2n+1+2n+1-21 2 a2n-2
=
=
(a2n-4n)+2n-11 2 a2n-2
=
a2n-11 2 a2n-2
(6分)1 2
∵b1=a2-2=-
(9分)1 2
∴数列{bn}是等比数列,且bn=-
×(1 2
)n-1=-(1 2
)n (l0分)1 2
(Ⅲ)由(Ⅱ)得a2n=bn+2=2-(
)n(n=1,2,…50)(12分)1 2
∴S=a2+a4+…+a100=2×50-
(1-1 2
)1 250 1- 1 2
=100-1+
=99+1 250
(14分)1 250