求数列（矩阵）的通项公式a(x,y)
1个回答
a(0,0)=1.
a(1,1)=a(0,0)+a(1,0)=1+0=1.
y>1时,
a(1,y)=ya(0,y-1)+a(1,y-1)=y+a(1,y-1)=(1/2)[y^2]-(1/2)(y-1)^2 + y/2 -(y-1)/2 + a(1,y-1)
a(1,y)-y^2/2-y/2 = a(1,y-1)-(y-1)^2/2-(y-1)/2 = ...= a(1,2-1) - (2-1)^2/2 - (2-1)/2
=a(1,1)-1/2-1/2=1-1=0
a(1,y)=y^2/2 - y/2=(y-1)y/2,y=2,3,...
a(1,1)=1,a(1,0)=0.
x>1时,
a(x,1)=a(x-1,0)+a(x,0)=0+0=0,x=2,3,...
a(1,1)=1,a(0,1)=1.
a(2,2)=2a(1,1)+a(2,1)=2a(1,1)=2
x>2时,
a(x,2)=2a(x-1,1)+a(x,1)=2a(x-1,1)=0,x=3,4,...
y>2时,
a(2,y)=ya(1,y-1)+a(2,y-1)=y[(y-1)^2/2-(y-1)/2] + a(2,y-1)=a(2,y-1)+(y-2)(y-1)y/2
=a(2,y-1) + (y-1)y(y+1)(y+2)/8 - (y-2)(y-1)y(y+1)/8
a(2,y)-(y-1)y(y+1)(y+2)/8 = a(2,y-1) - (y-1-1)(y-1)(y-1+1)(y-1+2)/8
=...=a(2,3-1)-(3-1-1)(3-1)(3-1+1)(3-1+2)/8
=a(2,2)- 3
=-1
a(2,y)=(y-1)y(y+1)(y+2)/8 - 1,y=2,3,4,...
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