由,知,故,再由n为偶数和n为奇数两种情况进行分类讨论,求和:.
【解析】
∵f(x)==,,
∴,
∵b1=1,∴{bn}是首项为1,公差为的等差数列,
∴,
∴,
①当n为偶数时:
∵,,
∴Wn=b1b2-b2b3+b3b4-b4b5+…+bn-1bn-bnbn+1
=b2(b1-b3)+b4(b3-b5)+…+bn(bn-1-bn+1)
=-2×(b2+b4+…+bn)
=-×
=-;
②当n为奇数时:
∵,,
∴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×+bnbn+1
=-×[]+
=.
故.