(x²-x+1)^n=a0+a1x+a2x²+...+a(2n)x^(2n) n∈N*,则a1+a2+a3+...+a(2n-1)=
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![(x²-x+1)^n=a0+a1x+a2x²+...+a(2n)x^(2n) n∈N*,则a1+a2+a3+...+a(2n-1)=](/uploads/image/z/11623682-2-2.jpg?t=%28x%26%23178%3B-x%2B1%29%5En%3Da0%2Ba1x%2Ba2x%26%23178%3B%2B...%2Ba%282n%29x%5E%282n%29+n%E2%88%88N%2A%2C%E5%88%99a1%2Ba2%2Ba3%2B...%2Ba%282n-1%29%3D)
(x²-x+1)^n=a0+a1x+a2x²+...+a(2n)x^(2n) n∈N*,则a1+a2+a3+...+a(2n-1)=
(x²-x+1)^n=a0+a1x+a2x²+...+a(2n)x^(2n) n∈N*,则a1+a2+a3+...+a(2n-1)=
(x²-x+1)^n=a0+a1x+a2x²+...+a(2n)x^(2n) n∈N*,则a1+a2+a3+...+a(2n-1)=
令x=0,则a0=1;令x=1,则a0+a1+a2+...+a2n=1;故a1+a2+...+a2n=0.(x^2-x+1)^n=(x^2+(1-x))^n=C(n,0)(x^2)^n+...,故a2n=C(n,0)=1,从而a1+a2+...+a(2n-1)=-1.