一道关于数列的题,数列{an}是非常数列,且满足a(n+1)+a(n-1)=2a(n),(n属于1,2,3.),设有函数

f(x)是1次多项式.
由a(n+1)+a(n-1)=2a(n)可知数列{an}是等差数列,设公差为d,
f(x)=a0*C(8,8)*(1-x)^8+a1*C(8,7)*x*(1-x)^7+a2*C(8,6)*x^2*(1-x)^6+a3*C(8,5)*x^3*(1-x)^5+...+a8*C(8,0)*x^8
=a0*C(8,8)*(1-x)^8+(a0+d)*C(8,7)*x*(1-x)^7+(a0+2d)*C(8,6)*x^2*(1-x)+...+(a0+8d)*C(8,0)*x^8
=a0(C(8,8)*(1-x)^8+C(8,7)*x*(1-x)^7+C(8,6)*x^2*(1-x)+...+C(8,0)*x^8)
+d(C(8,7)*x*(1-x)^7+2C(8,6)*x^2*(1-x)^6+3C(8,5)*x^3*(1-x)^5+...+8C(8,0)*x^8)
=a0((1-x)+x)^8+8xd(C(7,7)*(1-x)^7+C(7,6)*x*(1-x)^6+C(7,5)*x^2*(1-x)^5+...+C(7,0)*x^7) (这里用到组合恒等式kC(8,8-k)=8C(7,8-k))
=a0+8xd((1-x)+x)^7=a0+8xd,即
f(x)=8xd+a0.由d不等于零可知f(x)是1次多项式.