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)=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,求这个函数的次数.8次,7次,0
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![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)=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,求这个函数的次数.8次,7次,0](/uploads/image/z/4351209-33-9.jpg?t=%E4%B8%80%E9%81%93%E5%85%B3%E4%BA%8E%E6%95%B0%E5%88%97%E7%9A%84%E9%A2%98%2C%E6%95%B0%E5%88%97%7Ban%7D%E6%98%AF%E9%9D%9E%E5%B8%B8%E6%95%B0%E5%88%97%2C%E4%B8%94%E6%BB%A1%E8%B6%B3a%28n%2B1%29%2Ba%28n-1%29%3D2a%28n%29%2C%28n%E5%B1%9E%E4%BA%8E1%2C2%2C3.%29%2C%E8%AE%BE%E6%9C%89%E5%87%BD%E6%95%B0f%28x%29%3Da0%2AC%288%2C8%29%2A%281-x%29%5E8%2Ba1%2AC%288%2C7%29%2Ax%2A%281-x%29%5E7%2Ba2%2AC%288%2C6%29%2Ax%5E2%2A%281-x%29%5E6%2Ba3%2AC%288%2C5%29%2Ax%5E3%2A%281-x%29%5E5%2B...%2Ba8%2AC%288%2C0%29%2Ax%5E8%2C%E6%B1%82%E8%BF%99%E4%B8%AA%E5%87%BD%E6%95%B0%E7%9A%84%E6%AC%A1%E6%95%B0.8%E6%AC%A1%2C7%E6%AC%A1%2C0)
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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)=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,求这个函数的次数.8次,7次,0
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)=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,求这个函数的次数.8次,7次,0次,1次.
C(8,7)为排列组合那块的组合.
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)=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,求这个函数的次数.8次,7次,0
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.