15.数列{an}满足:a1 = 1,且对任意的 m,n属于正自然数,都有am+n(m+n为下标)=am+an+nm则1/a1+1/a2+1/a3.+1/an=?答案为2n/(n+1)
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![15.数列{an}满足:a1 = 1,且对任意的 m,n属于正自然数,都有am+n(m+n为下标)=am+an+nm则1/a1+1/a2+1/a3.+1/an=?答案为2n/(n+1)](/uploads/image/z/7125254-62-4.jpg?t=15.%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3%EF%BC%9Aa1+%3D+1%2C%E4%B8%94%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84+m%2Cn%E5%B1%9E%E4%BA%8E%E6%AD%A3%E8%87%AA%E7%84%B6%E6%95%B0%2C%E9%83%BD%E6%9C%89am%2Bn%EF%BC%88m%2Bn%E4%B8%BA%E4%B8%8B%E6%A0%87%EF%BC%89%3Dam%2Ban%2Bnm%E5%88%991%2Fa1%2B1%2Fa2%2B1%2Fa3.%2B1%2Fan%3D%3F%E7%AD%94%E6%A1%88%E4%B8%BA2n%2F%28n%2B1%29)
15.数列{an}满足:a1 = 1,且对任意的 m,n属于正自然数,都有am+n(m+n为下标)=am+an+nm则1/a1+1/a2+1/a3.+1/an=?答案为2n/(n+1)
15.数列{an}满足:a1 = 1,且对任意的 m,n属于正自然数,都有am+n(m+n为下标)=am+an+nm
则1/a1+1/a2+1/a3.+1/an=?
答案为2n/(n+1)
15.数列{an}满足:a1 = 1,且对任意的 m,n属于正自然数,都有am+n(m+n为下标)=am+an+nm则1/a1+1/a2+1/a3.+1/an=?答案为2n/(n+1)
令m=1
则a(n+1)=an+1+n
a(n+1)-an=n+1
所以
an-a(n-1)=n
a(n-1)-a(n-2)=n-1
……
a2-a1=2
相加
an-a1=2+3+……+n
a1=1
an=1+2+3+……+n=n(n+1)/2
1/an=2/n(n+1)=2[1/n-1/(n+1)]
所以原式=2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)
令m=1 得an+1=an+n+1 得an=n(n+1)/2 (累加法)
原式=2*(1-1/2+1/2-1/3+1/3-…+1/n-1/(n+1))=2n/(n+1) (裂项求和)
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