12/02/17 18:31:57.94 0g295DUg0
>>15
m=1
(n+2)a[n]-(2n+1)a[1]+(1-n)a[n+1]=0
a[1]=0
a[n+1]=(n+2)/(n-1)a[n]
a[n]=(n+1)/(n-2)a[n-1]=(n+1)n(n-1)/(3・2・1)a[2]=(n+1)n(n-1)
(n+2m)(n+1)n(n-1)-(m+2n)(m+1)m(m-1)+(m-n)(n+m+1)(n+m)(n+m-1)=0
a[1]=1
(n-1)a[n+1]=(n+2)a[n]-(2n+1)
(n-1)(a[n+1]-(n+1))=(n+2)(a[n]-n)
a[n]-n=(n+1)n(n-1)/(3・2・1)(a[2]-2)=0
a[n]=n
(n+2m)n-(m+2n)m+(m-n)(m+n)=0
(a[1],a[2])=(a[2]/6-a[1]/3)(0,6)+a[1](1,2)
a[n]=(a[2]/6-a[1]/3)(n+1)n(n-1)+a[1]n=((n+1)n(n-1)/6)a[2]-((n+2)n(n-2)/3)a[1]