已知递推公式,求通项公式(高分求..)

Question

如题所述

Answer 1

a[i]=n*a[i-1]+(n+1)*a[i-2]
所以a[i+1]=n*a[i]+(n+1)*a[i-1]
所以a[i+1]+a[i]=(n+1)a[i]+(n+1)a[i-1]=(n+1)(a[i]+a[i-1])
设a[i+1]+a[i]=b[i]
则有
b[i]=(n+1)b[i-1]
所以b[i]是等比数列
b[i]=b1*(n+1)^(i-1)
b1=a2+a1
所以b[i]=a[i+1]+a[i]=(a1+a2)*(n+1)^(i-1)
所以:
a2+a1=a1+a2
a3+a2=(a1+a2)*(n+1)
a4+a3=(a1+a2)*(n+1)^2
......
......
当i为偶数时:
a2+a1=(a1+a2)*[-(n+1)]^0
-a3-a2=(a1+a2)*[-(n+1)]^1
a4+a3=(a1+a2)*[-(n+1)]^2
......
......
a[i]+a[i-1]=(a1+a2)*[-(n+1)]^(i-2)
各式叠加,有
a[i]+a1=(a1+a2)*[1-(-(n+1))^(i-1)]/[1-(-(n+1))]=(a1+a2)*[1+(n+1)^(i-1)]/(n+2)
所以,a[i]=(a1+a2)*[(n+1)^(i-1)+1]/(n+2)
-
a1
当i为奇数时:
a2+a1=(a1+a2)*[-(n+1)]^0
-a3-a2=(a1+a2)*[-(n+1)]^1
a4+a3=(a1+a2)*[-(n+1)]^2
......
......
-a[i]-a[i-1]=(a1+a2)*[-(n+1)]^(i-2)
各式相加得:
-a[i]+a1=(a1+a2)*[1-(n+1)^(i-1)]/(n+2)
a[i]=(a1+a2)*[(n+1)^(i-1)-1]/(n+2)
+
a1

Answer 2

an=na(n-1)+(n+1)a(n-2)么？变形得an+a(n-1)=(n+1)[a(n-1)+a(n-2)].递推知an+a(n-1)=(n+1)!(a2+a1)/3!.令bn=an+a(n-1)则an=∑[(-1)^n×bn]=(a2+a1)/6×∑(n×n!)只能化到这里了。