已知数列{an}满足a1=1,an=a(n-1)+1/n(n-1)[n>=2],则通项公式an.

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已知数列{an}满足a1=1,an=a(n-1)+1/n(n-1)[n>=2],则通项公式an.已知数列{an}满足a1=1,an=a(n-1)+1/n(n-1)[n>=2],则通项公式an.已知数列

已知数列{an}满足a1=1,an=a(n-1)+1/n(n-1)[n>=2],则通项公式an.
已知数列{an}满足a1=1,an=a(n-1)+1/n(n-1)[n>=2],则通项公式an.

已知数列{an}满足a1=1,an=a(n-1)+1/n(n-1)[n>=2],则通项公式an.
原式可化为 an=a(n-1)+1/(n-1)-1/n
an+1/n=a(n-1)+1/(n-1)
an+1/n=a1+1/1=2
所以an=2-1/n

原式可化为 an=a(n-1)+1/(n-1)-1/n
an+1/n=a(n-1)+1/(n-1)
an+1/n=a1+1/1=2
所以an=2-1/n

an=a(n-1)+1/(n-1)-1/n
a(n-1)=a(n-2)+1/(n-2)(n-1)
....
a2=a1+1/2
此时n>1
得到an=a1+1/(1*2)+1/(2*3)+......+1/(n-1)n
而对于1/(n-1)n=[1/(n-1)]-1/n，那么上式化成
an=1+1-1/2+1/2-1/3+1/3-1/4+.....+1/(n-1)-1/n=2-1/n
验证，当n=1时a1=2-1/1=1
故an=2-1/n (n>=1)

an=a(n-1)+[(1/(n-1))-(1/n)]
an+(1/n)=a(n-1)+(1/(n-1))
由于a1+(1/1)=2，则an+(1/n)是以2为首项，0为公差的等差数列，
或以2为首项，1为公比的等比数列，
则an+(1/n)=2，则an=2-(1/n)

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