等差数列an中,公差d>0,ana(n+1)=4n^2-1,求通项公式an

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等差数列an中,公差d>0,ana(n+1)=4n^2-1,求通项公式an
等差数列an中,公差d>0,ana(n+1)=4n^2-1,求通项公式an

等差数列an中,公差d>0,ana(n+1)=4n^2-1,求通项公式an
设A1=a 公差=d
An=a+(n-1)d=a-d+nd A(n+1)=a+nd
AnA(n+1)=(a-d+nd)(a+nd)
=(nd)^2+(2a-d)nd+a^2+a(a-d)
=4n^2-1
d^2=4 (2a-d)d=0 a(a-d)=-1
d=2 a=1
An=2n-1

an=4n-1

an=4n-4

2n-1 肯定对