分析:(1)把
a1=代入a
n+1=|a
n-1|分別求得a
2,a
3,a
4,推斷出n≥2數(shù)列中偶數(shù)項(xiàng)為
,奇數(shù)項(xiàng)為
,進(jìn)而推斷出數(shù)列的通項(xiàng)公式.
(2)根據(jù)a
1=a可分別求得a
2和a
3,同理可求得a
k+1,a
k+2,a
k+3,a
k+4進(jìn)而求得a
3k和a
3k-1最后相加,利用等差數(shù)列的求和公式求得答案.
解答:解:(1)
a1=,a2=,a3=,a4=,
∴
a1=,n≥2時,
an=,其中k∈N
*
(2)當(dāng)a
1=a∈(k,k+1),(k∈N
*)時,
易知a
2=a-1,
a
3=a-2a
k=a-(k-1);
a
k+1=a-k∈(0,1);
a
k+2=1-a
k+1=k+1-a;
a
k+3=1-a
k+2=a-k;
a
k+4=1-a
k+3=k+1-a
a
3k-1=a-k,
a
3k=k+1-a;
S
3k=a
1+a
2+…+a
k+a
k+1+a
k+2+a
k+3+a
k+4+…+a
3k-1+a
3k
=a+(a-1)+(a-2)+…+a-(k-1)+k
=
ka+k-(k-1)=
-+k(a+) 點(diǎn)評:本題主要考查了數(shù)列的遞推式.考查了學(xué)生推理分析和基本的運(yùn)算能力.