问题
解答题
设函数f(x)=
求和:Wn=b1b2-b2b3+b3b4-…+(-1)n-1•bnbn+1. |
答案
∵f(x)=
=2x+3 3x
+2 3
,bn=f(1 x
),n≥2,1 bn-1
∴bn=
+bn-1,2 3
∵b1=1,∴{bn}是首项为1,公差为
的等差数列,2 3
∴bn=
,2n+1 3
∴bnbn+1=
(4n2+8n+3),1 9
①当n为偶数时:
∵b2=
,bn=5 3
,2n+1 3
∴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 3
=-
×[4 3
(n 4
+5 3
)]2n+1 3
=-
(2n2+6n);1 9
②当n为奇数时:
∵b2=
,bn-1=5 3
,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×
(b2+b4+…+bn-1)+bnbn+12 3
=-
×[4 3
(n-1 4
+5 3
)]+2n-1 3
(4n2+8n+3)1 9
=
(2n2+6n+7).1 9
故Wn=
.-
(2n2+6n),n为偶数1 9
(2n2+6n+7),n为奇数1 9