A_{n}+A_{n+1}-1=n*(A_{n+1}-A_{n-1})-------------------------1
A_{n-1}+A_{n}-1=(n-1)*(A_{n}-A_{n-2})-------------------------2
用1-2,得:
A_{n+1}-A_{n-1}=n*(A_{n+1}-A_{n-1})-(n-1)*(A_{n}-A_{n-2});
(n-1)*(A_{n+1}-A_{n-1})=(n-1)*(A_{n}-A_{n-2});
1) n=1,代入 1 式,
A_{1}=1;
2) n/=1,有递推关系:
A_{n+1}-A_{n-1}=A_{n}-A_{n-2}=d;
可见数列为隔项等差数列.公差为d
n=2,有 A_{2}=d.
所以 通项为:
A_{2n}=n*A_{2};
A_{2n+1}=n*A_{2}+1; (n为自然数)