如果数列满足a1=1,an-n+3=an-1 【a的n-1】（n大于等于2）,求an

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如果数列满足a1=1,an-n+3=an-1【a的n-1】（n大于等于2）,求an如果数列满足a1=1,an-n+3=an-1【a的n-1】（n大于等于2）,求an如果数列满足a1=1,an-n+3=

如果数列满足a1=1,an-n+3=an-1 【a的n-1】（n大于等于2）,求an
如果数列满足a1=1,an-n+3=an-1 【a的n-1】（n大于等于2）,求an

如果数列满足a1=1,an-n+3=an-1 【a的n-1】（n大于等于2）,求an
a2-2+3=a1,那么a2=0,a3=0
an-n+3=an-1,
an-an-1=n-3(n>=2)
.
a2-a1=2-3
相加得an-a1=2+3+.+n-3*（n-1）
所以a>1时,an=(n²-5n+6)/2
n=1时,代入上式a1=1,即n=1时也成立
所以an=(n²-5n+6)/2,n=1,2,3.

an=2+（n-2）（n-3）/2

an-a(n-1)=n-3(n≥2)
假设数列bn=n-3 （n≥2)
bn的前n-1项和 S（n-1）=b2+b3+b4+ +bn
=(n-1)[(2-3)+(n-3)]/2
=n^2/2-5n/2+2(n≥2)
另一方面：...

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an-a(n-1)=n-3(n≥2)
假设数列bn=n-3 （n≥2)
bn的前n-1项和 S（n-1）=b2+b3+b4+ +bn
=(n-1)[(2-3)+(n-3)]/2
=n^2/2-5n/2+2(n≥2)
另一方面： S（n-1）=(a2-a1)+(a3-a2)+(a4-a3)+ +[a(n-1)-a(n-2)]+[an-a(n-1)]
=an-a1(n≥2)
所以：an=Sn+a1
=n^2/2-5n/2+2+1
=n^2/2-5n/2+3(n≥2)
n=1时满足上式。所以an==n^2/2-5n/2+3(n≥1)

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