用完全归纳法证明递归,递归前提:a_0=0,a_1=1,a_(n+1)=a_n+a_(n-1) n≥1

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用完全归纳法证明递归,递归前提:a_0=0,a_1=1,a_(n+1)=a_n+a_(n-1) n≥1
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用完全归纳法证明递归,递归前提:a_0=0,a_1=1,a_(n+1)=a_n+a_(n-1) n≥1
用完全归纳法证明递归,
递归前提:a_0=0,a_1=1,a_(n+1)=a_n+a_(n-1) n≥1

用完全归纳法证明递归,递归前提:a_0=0,a_1=1,a_(n+1)=a_n+a_(n-1) n≥1
a) ① a_2=1+0=1
a_0*a_2-(a_1)^2=-1=(-1)^1 成立
②设n=k时成立
当n=k+1时 a_k*a_(k+2)-[a_(k+1)]^2=a_k*[a_k+a_(k+1)]-[a_(k+1)]^2=(a_k)^2+a_k*a_(k+1))-[a_(k+1)]^2=(a_k)^2+a_(k+1)*[a_k-a_(k+1)]=(a_k)^2-a_(k+1)*a_(k-1)=(-1)^(k+1)
成立
由①②知a)等式成立