在数列an中,a1=1/2,点(an,an+1)在直线y=x+1/2上.记bn=1/anan+1,求数列bn的前n项和Tn
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![在数列an中,a1=1/2,点(an,an+1)在直线y=x+1/2上.记bn=1/anan+1,求数列bn的前n项和Tn](/uploads/image/z/12680228-20-8.jpg?t=%E5%9C%A8%E6%95%B0%E5%88%97an%E4%B8%AD%2Ca1%3D1%2F2%2C%E7%82%B9%28an%2Can%2B1%29%E5%9C%A8%E7%9B%B4%E7%BA%BFy%3Dx%2B1%2F2%E4%B8%8A.%E8%AE%B0bn%3D1%2Fanan%2B1%2C%E6%B1%82%E6%95%B0%E5%88%97bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn)
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在数列an中,a1=1/2,点(an,an+1)在直线y=x+1/2上.记bn=1/anan+1,求数列bn的前n项和Tn
在数列an中,a1=1/2,点(an,an+1)在直线y=x+1/2上.记bn=1/anan+1,求数列bn的前n项和Tn
在数列an中,a1=1/2,点(an,an+1)在直线y=x+1/2上.记bn=1/anan+1,求数列bn的前n项和Tn
你确定是bn=1/anan+1,而不是bn=1/an +1 ?
点(an,an+1)在直线y=x+1/2上
a(n+1) = an + 1/2
a(n+1) -an = 1/2
an - a1 = (n-1)/2
an = n
bn=1/(an.a(n+1))
= 1/[n(n+1)]
=1/n -1/(n+1)
Tn =b1+b2+...+bn
=1- 1/(n+1)
=n/(n+1)