“離散的”な世界数学的帰納法と数列数列の帰納的定義漸化式、階差数列、一次式の解、推定、等差数列、等比級数

学習環境

新装版数学読本3 (松坂和夫(著)、岩波書店)の第13章(“離散的”な世界 - 数列)、13.2(数学的帰納法と数列)、数列の帰納的定義の問32の解答を求めてみる。

階差数列。

b_{n} = a_{n + 1} - a_{n} = 2 n

よって、

\begin{array}{l} n \geq 2 \\ a_{n} = a_{1} + \sum_{k = 1}^{n - 1} b_{n} = 1 + 2 \cdot \frac{n (n - 1)}{2} = n^{2} - n + 1 \end{array}

また、

1^{2} - 1 + 1 = 1 = a_{n}

よって、求める一般項は

a_{n} = n^{2} - n + 1

\begin{array}{l} a_{n + 1} - a_{n} = 3^{n} \\ n \geq 2 \\ a_{n} = a_{1} + \sum_{k = 1}^{n - 1} 3^{k} = 2 + \frac{3 (3^{n - 1} - 1)}{3 - 1} = \frac{3^{n} + 1}{2} \\ \frac{3^{1} + 1}{2} = 2 = a_{1} \end{array}

一般頃。

a_{n} = \frac{3^{n} + 1}{2}

\begin{array}{l} x = 3 x - 2 \\ x = 1 \end{array}

\begin{array}{l} a_{n + 1} - 1 = 3 (a_{n} - 1) \\ a_{1} - 1 = 2 - 1 = 1 \\ a_{n} - 1 = 3^{n - 1} \\ a_{n} = 3^{n - 1} + 1 \end{array}

\begin{array}{l} x = - 2 x + 9 \\ x = 3 \end{array}

\begin{array}{l} a_{n + 1} - 3 = - 2 (a_{n} - 3) \\ a_{1} - 3 = 1 - 3 = - 2 \\ a_{n} - 3 = (- 2) {(- 2)}^{n - 1} = {(- 2)}^{n} \\ a_{n} = {(- 2)}^{n} + 3 \end{array}

\begin{array}{l} 2 x = x - 6 \\ x = - 6 \end{array}

\begin{array}{l} 2 (a_{n + 1} + 6) = a_{n} + 6 \\ a_{n + 1} + 6 = \frac{1}{2} (a_{n} + 6) \\ a_{1} + 6 = 2 + 6 = 8 \\ a_{n} + 6 = 8 \cdot {(\frac{1}{2})}^{n - 1} \\ a_{n} = \frac{8}{2^{n - 1}} - 6 \end{array}

\begin{array}{l} a_{1} = 1 \\ a_{2} = \frac{1}{2} \\ a_{3} = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3} \end{array}

a_{n + 1} = \frac{n}{n + 1} \cdot \frac{1}{n} = \frac{1}{n + 1}

よって、帰納法により

a_{n} = \frac{1}{n}

\begin{array}{l} a_{1} = 2 \\ a_{2} = \frac{2}{2 + 1} = \frac{2}{3} \\ a_{3} = \frac{\frac{2}{3}}{\frac{2}{3} + 1} = \frac{2}{3} \cdot \frac{3}{5} = \frac{2}{5} \\ a_{4} = \frac{\frac{2}{5}}{\frac{2}{5} + 1} = \frac{2}{5} \cdot \frac{5}{7} = \frac{2}{7} \\ a_{5} = \frac{\frac{2}{7}}{\frac{2}{7} + 1} = \frac{2}{7} \cdot \frac{7}{9} = \frac{2}{9} \end{array}

\begin{array}{l} \frac{2}{2 \cdot 1 - 1} = 2 = a_{1} \\ a_{n + 1} = \frac{a_{n}}{a_{n + 1} + 1} = \frac{\frac{2}{2 n - 1}}{\frac{2}{2 n - 1} + 1} = \frac{2}{2 n - 1} \cdot \frac{2 n - 1}{2 n + 1} = \frac{2}{2 (n + 1) - 1} \end{array}

よって、帰納法により、

a_{n} = \frac{2}{2 n - 1}

\begin{array}{l} a_{1} = 3 \\ a_{2} = 3 \\ a_{3} = 6 \\ a_{4} = 12 \\ a_{5} = 24 \end{array}

a_{2} = 3 = 3 \cdot 2^{n - 2}

a_{n + 1} = \sum_{k = 1}^{n} a_{k}

= 3 + \sum_{k = 2}^{n} a_{k}

= 3 + \sum_{k = 2}^{n} 3 \cdot 2^{k - 2}

= 3 + 3 \cdot \frac{2^{n - 1} - 1}{2 - 1}

= 3 + 3 (2^{n - 1} - 1)

= 3 \cdot 2^{n - 1}

= 3 \cdot 2^{{(n + 1)}^{-} - 2}

よって、帰納法より一般項は、

\begin{array}{l} a_{1} = 3 \\ n \geq 2 \\ a_{n} = 3 \cdot 2^{n - 2} \end{array}

コード(Wolfram Language, Jupyter)

a1[n_] := n^2-n+1

a1[1] == 1

a1[n + 1] == a1[n] + 2 n

Simplify[%]

a2[n_] := (3^n+1)/2

a2[1] == 2

a2[n+1] == a2[n] + 3^n

Simplify[%]

a3[n_] := 3^(n-1) + 1

a3[1] == 2

Simplify[a3[k + 1] == 3a3[k] - 2]

a4[n_] := (-2)^n+3

a4[1] == 1

Simplify[a4[k + 1] == -2a4[k] + 9]

a[n_] := 8/2^(n-1) - 6

a[1] == 2

Simplify[2a[k+1] == a[k] - 6]

a[n_] := 1/n

a[1] == 1

Simplify[a[k+1] == k/(k + 1)a[k]]

a[n_] := 2/(2n-1)

a[1] == 2

a[k+1] == a[k] / (a[k] + 1)

Simplify[%]

a[n_] := 3 2^(n-2)

a[k+1] == 3 + Sum[a[i], {i, 2, k}]

Simplify[%]