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幼苗
共回答了17个问题采纳率:82.4% 举报
∵f(x)=[2x+3/3x]=[2/3+
1
x],bn=f(
1
bn−1),n≥2,
∴bn=
2
3+bn−1,
∵b1=1,∴{bn}是首项为1,公差为[2/3]的等差数列,
∴bn=
2n+1
3,
∴bnbn+1=
1
9(4n2+8n+3),
①当n为偶数时:
∵b2=
5
3,bn=
2n+1
3,
∴Wn=b1b2-b2b3+b3b4-b4b5+…+bn-1bn-bnbn+1
=b2(b1-b3)+b4(b3-b5)+…+bn(bn-1-bn+1)
=-2×[2/3](b2+b4+…+bn)
=-[4/3]×[
n
4(
5
3+
2n+1
3)]
=-[1/9(2n2+6n);
②当n为奇数时:
∵b2=
5
3],bn−1=
2n−1
3,
∴Wn=b1b2-b2b3+b3b4-b4b5+…+bn-2bn-1-bn-1bn+bnbn+1
=b2(b1-b3)+b4(b3-b5)+…+bn-1(bn-2-bn)+bnbn+1
=-2×[2/3(b2+b4+…+bn−1)+bnbn+1
=-
4
3]×[
n−1
4(
5
3+
2n−1
3)]+
1年前
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