Określ rekurencyjnie ciąg ...- Zadanie Ćwiczenie 3: MATeMAtyka 2. Zakres rozszerzony. Po gimnazjum

Rozwiązanie

$a)$

$a_{n} = 2 n + 1$

$a_{1} = 2 + 1 = 3$

$a_{n + 1} = 2 n + 3$

$a_{n + 1} - a_{n} = 2 n + 3 - 2 n - 1 = 2$

${a_{1} = 3 a_{n + 1} = a_{n} + 2 n \geq 1$

$lub :$

$\frac{a _{n} + 1}{a _{n}} = \frac{2 n + 3}{2 n + 1}$

${a_{1} = 3 a_{n + 1} = \frac{2 n + 3}{2 n + 1} \cdot a_{n} n \geq 1$

$b)$

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

$a_{1} = 1 + 1 = 2$

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

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

${a_{1} = 2 a_{n + 1} = a_{n} + 2 n + 1 n \geq 1$

$lub:$

$\frac{a _{n} + 1}{a _{n}} = \frac{n ^{2} + 2 n + 2}{n ^{2} + 1}$

${a_{1} = 2 a_{n + 1} = \frac{n ^{2} + 2 n + 2}{n ^{2} + 1} \cdot a_{n} n \geq 1$

$c)$

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

$a_{1} = 2 \cdot 3 = 6$

$a_{n + 1} = 2 \cdot 3^{n + 1} = 6 \cdot 3^{n}$

$a_{n + 1} - a_{n} = 6 \cdot 3^{n} - 2 \cdot 3^{n} = 3^{n} (6 - 2) = 3^{n} \cdot 4$

${a_{1} = 6 a_{n + 1} = a_{n} + 3^{n} \cdot 4 n \geq 1$

$lub:$

$\frac{a _{n + 1}}{a _{n}} = \frac{6 \cdot 3 ^{n}}{2 \cdot 3 ^{n}} = 3$

${a_{1} = 6 a_{n + 1} = 3 a_{n} n \geq 1$

$d)$

$a_{n} = (- 1)^{n}$

$a_{1} = - 1$

$a_{n + 1} = (- 1)^{n + 1}$

$a_{n + 1} - a_{n} = (- 1)^{n + 1} - (- 1)^{n} = (- 1)^{n} (- 1 - 1) = - 2 (- 1)^{n}$

${a_{1} = - 1 a_{n + 1} = a_{n} - 2 (- 1)^{n} n \geq 1$

$lub:$

$\frac{a _{n + 1}}{a _{n}} = \frac{( - 1 ) ^{n + 1}}{( - 1 ) ^{n}} = - 1$

${a_{1} = - 1 a_{n + 1} = - a_{n} n \geq 1$

$e)$

$a_{n} = \frac{n ( n + 1 )}{2}$

$a_{1} = 1$

$a_{n + 1} = \frac{( n + 1 ) ( n + 2 )}{2}$

$a_{n + 1} - a_{n} = \frac{( n + 1 ) ( n + 2 )}{2} - \frac{n ( n + 1 )}{2} = \frac{( n + 1 ) ( n + 2 - n )}{2} = \frac{( n + 1 ) 2}{2} = (n + 1)$

${a_{1} = 1 a_{n + 1} = a_{n} + n + 1 n \geq 1$

$lub:$

$\frac{a _{n + 1}}{a _{n}} = \frac{\frac{( n + 1 ) ( n + 2 )}{2}}{\frac{n ( n + 1 )}{2}} = \frac{n + 2}{n}$

${a_{1} = 1 a_{n + 1} = \frac{n + 2}{2} a_{n} n \geq 1$

$f)$

$a_{n} = n + 1$

$a_{1} = 2$

$a_{n + 1} = n + 2$

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

${a_{1} = 2 a_{n + 1} = a_{n} + n + 2 - n + 1 n \geq 1$

$lub:$

$\frac{a _{n + 1}}{a _{n}} = \frac{n + 2}{n + 1}$

$⎩ ⎨ ⎧ a_{1} = 2 a_{n + 1} = \frac{n + 2}{n + 1} a_{n} n \geq 1$

Krystian

Nauczyciel matematyki

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