设f(x)=ax^2(x的2平方）+bx+c,证明：f(x+3)-3f(x+2)+3f(x+1)-f(x)=0请帮我解答一下吧!

Question

设f(x)=ax^2(x的2平方）+bx+c,证明：f(x+3)-3f(x+2)+3f(x+1)-f(x)=0
请帮我解答一下吧!

Answer 1

f(x+3)-3f(x+2)+3f(x+1)-f(x)=a(x+3)^2+b(x+3)-3[a(x+2)^2+b(x+2)]+3[a(x+1)^2+b(x+1)]-ax^2-bx=a[(x+3)^2-3(x+2)^2+3(x+1)^2-x^2]+b[(x+3)-3(x+2)+3(x+1)-x] (x+3)^2-3(x+2)^2+3(x+1)^2-x^2=0 (x+3)-3(x+2)+3(x+1)-x=0 所以f(x+3)-3f(x+2)+3f(x+1)-f(x)=0