A1=0,aN+1=an+2n-1,求数列an的通项公式
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A1=0,aN+1=an+2n-1,求数列an的通项公式A1=0,aN+1=an+2n-1,求数列an的通项公式A1=0,aN+1=an+2n-1,求数列an的通项公式∵a1=0a(n+1)=an+2
A1=0,aN+1=an+2n-1,求数列an的通项公式
A1=0,aN+1=an+2n-1,求数列an的通项公式
A1=0,aN+1=an+2n-1,求数列an的通项公式
∵a1=0 a(n+1)=an+2n-1
∴a(n+1)-an=2n-1
∴ an-a(n-1)=2(n-1)-1
a(n-1)-a(n-2)=2(n-2)-1
.
a2-a1=2*1-1
将上面n-1个式子加起来得:
an-a1=2*1+2*2+...+2*(n-2)+2*(n-1)-(1+1+...+1)
=2*[1+2+.+(n-1)]-(n-1)
=2*(n-1)(1+n-1)/2 -(n-1)
=n(n-1)-(n-1)
=(n-1)^2
又∵a1=0
∴an=(n-1)^2
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