数列{an}满足a1=1,an=a(n-1)+1/(n2-n),求数列的通项公式
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数列{an}满足a1=1,an=a(n-1)+1/(n2-n),求数列的通项公式
数列{an}满足a1=1,an=a(n-1)+1/(n2-n),求数列的通项公式
数列{an}满足a1=1,an=a(n-1)+1/(n2-n),求数列的通项公式
∵n≥2,an=a(n-1)+1/(n2-n)
an-a(n-1)=1/[(n-1)n]=1/(n-1)-1/n
a2-a1=1-1/2
a3-a2=1/2-1/3
a4-a3=1/3-1/4
.
an-a(n-1)=1/(n-1)-1/n
将上述n-1个式子两边相加
an-a1=1-1/n=
an=2-1/n
n=1时,上式成立
∴an=2-1/n
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