證明16的倍數和費波那契數列問題

證明16的倍數和費波那契數列問題

1.
$$\forall n \in N$$，$$n$$為奇數，利用數學歸納法證明$$(n^2+6n-3)(n+3)$$恆為16的倍數。

2.

(1)試證：$$a_{n+2}=a_{n+1}+a_n$$，$$\forall n \in N$$恆成立
(2)求$$a_{10}=$$？
(3)試證：$$a_n \in N$$,$$\forall n \in N$$恆成立

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回復 1# thankyou 的帖子

\begin{align} & {{a}_{n+1}}+{{a}_{n}} \\ & =\frac{1}{\sqrt{5}}\left[ {{\left( \frac{1+\sqrt{5}}{2} \right)}^{n+1}}-{{\left( \frac{1-\sqrt{5}}{2} \right)}^{n+1}}+{{\left( \frac{1+\sqrt{5}}{2} \right)}^{n}}-{{\left( \frac{1-\sqrt{5}}{2} \right)}^{n}} \right] \\ & =\frac{1}{\sqrt{5}}\left[ {{\left( \frac{1+\sqrt{5}}{2} \right)}^{n}}\left( \frac{1+\sqrt{5}}{2}+1 \right)-{{\left( \frac{1-\sqrt{5}}{2} \right)}^{n}}\left( \frac{1-\sqrt{5}}{2}+1 \right) \right] \\ & =\frac{1}{\sqrt{5}}\left[ {{\left( \frac{1+\sqrt{5}}{2} \right)}^{n}}{{\left( \frac{1+\sqrt{5}}{2} \right)}^{2}}-{{\left( \frac{1-\sqrt{5}}{2} \right)}^{n}}{{\left( \frac{1-\sqrt{5}}{2} \right)}^{2}} \right] \\ & =\frac{1}{\sqrt{5}}\left[ {{\left( \frac{1+\sqrt{5}}{2} \right)}^{n+2}}-{{\left( \frac{1-\sqrt{5}}{2} \right)}^{n+2}} \right] \\ & ={{a}_{n+2}} \\ \end{align}

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回復 1# thankyou 的帖子

1.

(n^2+6n-3)(n+3) = [(2m-1)^2+6(2m-1)-3][(2m-1)+3] = [4m^2+8m-8][2m+2] = 8(m^2+2m-2)(m+1) = 8(m^3+3m^2-2)，所以只要再證明(m^3+3m^2-2)恆為偶數即可。

(k+1)^3+3(k+1)^2-2
≡k^3+6k^2+9k+2
≡k^3+k
≡-k+k.....將甲式代入
≡0 (mod 2)

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