数列求和 如果an>0,a1=1,an^2-(an-1)^2=(an-1)*an,如果an>0,a1=1,an^2-(an-1)^2=(an-1)*an,那么1/(a1+a2) +1/(a2+a3) +……+1/(an-1+an) =
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数列求和 如果an>0,a1=1,an^2-(an-1)^2=(an-1)*an,如果an>0,a1=1,an^2-(an-1)^2=(an-1)*an,那么1/(a1+a2) +1/(a2+a3) +……+1/(an-1+an) =
数列求和 如果an>0,a1=1,an^2-(an-1)^2=(an-1)*an,
如果an>0,a1=1,an^2-(an-1)^2=(an-1)*an,那么1/(a1+a2) +1/(a2+a3) +……+1/(an-1+an) =
数列求和 如果an>0,a1=1,an^2-(an-1)^2=(an-1)*an,如果an>0,a1=1,an^2-(an-1)^2=(an-1)*an,那么1/(a1+a2) +1/(a2+a3) +……+1/(an-1+an) =
n≥2时,
an²-a(n-1)²=a(n-1)an
an²-ana(n-1)=a(n-1)²
等式两边同除以a(n-1)²
[an/a(n-1)]²-[an/a(n-1)]=1
[an/a(n-1) -1/2]²=5/4
an/a(n-1) -1/2=√5/2或an/a(n-1) -1/2=-√5/2(舍去)
an/a(n-1)=(1+√5)/2,为定值.
a1=1,数列{an}是以1为首项,(1+√5)/2为公比的等比数列
an=1×[(1+√5)/2]^(n-1)=[(1+√5)/2]^(n-1)
1/an=[2/(1+√5)]^(n-1)=[(√5-1)/2]^(n-1)
1/[an²-a(n-1)²]=1/[ana(n-1)]
1/[(an+a(n-1))(an-a(n-1))]=1/[ana(n-1)]
1/[an+a(n-1)]=[an-a(n-1)]/[ana(n-1)]=1/a(n-1)-1/an
1/(a1+a2)+1/(a2+a3)+...+1/[a(n-1)+an]
=1/a1-1/a2+1/a2-1/a3+...+1/a(n-1)-1/an
=1/a1-1/an
=1/1 -[(√5-1)/2]^(n-1)
=1- [(√5-1)/2]^(n-1)
解;设an=m,an-1=n
根据题意得;m^2-n^2=mn
即(m+n)(m-n)=mn
所以1/(m+n)=(m-n)/mn=1/n-1/m
即1/(an-1+an)=1/(an-1)-1/an
所以1/(a1+a2) +1/(a2+a3) +……+1/(an-1+an)
=1/a1-1/a2+1/a2-1/a3+……+1/an-1-1/an
=1/a1-1/an
=1-1/an
后面就不写了 1L正解。