由 Raceleader 於 星期六 八月 02, 2003 11:33 am
f(x)+f(x+2)=2f(x+1)+2x
f(x)-f(x+1)=f(x+1)-f(x+2)+2x
f(0)-f(1)=f(1)-f(2)+0 ---(1)
f(1)-f(2)=f(2)-f(3)+2 ---(2)
f(2)-f(3)=f(3)-f(4)+4 ---(3)
...
f(n)-f(n+1)=f(n+1)-f(n+2)+2n ---(n+1)
(1)+(2)+(3)+...+(n+1):
f(0)-f(n+1)=f(1)-f(n+2)+n(n+1)
1-f(n+1)=0-f(n+2)+n(n+1)
f(n+2)-f(n+1)=n(n+1)-1
f(2)-f(1)=-1 ---(1)
f(3)-f(2)=1 ---(2)
f(4)-f(3)=5 ---(3)
...
f(n)-f(n-1)=(n-2)(n-1)-1 ---(n-1)
(1)+(2)+(3)+...+(n-1):
f(n)-f(1)=(1/6)(n-1)(n-2)(2n-3)+(1/2)(n-1)(n-2)+1
f(n)-0=(1/6)(n-1)(n-2)(2n-3)+(1/2)(n-1)(n-2)+1
f(x)=(1/6)(x-1)(x-2)(2x-3)+(1/2)(x-1)(x-2)+1
f(x)=x3/3-x2+(2/3)x+1