Odpowiedź:
a)
an = (n + 1)² ; n ∈ N⁺
(n + 1)² = 64
n² + 2n + 1 = 64
n² + 2n + 1 - 64 = 0
n² + 2n - 63 = 0
a = 1 , b = 2 , c = - 63
Δ = b² - 4ac = 2² - 4 * 1 * (- 63) = 4 + 252 = 256
√Δ = √256 = 16
n₁ = (- b - √Δ)/2a = (- 2 - 16)/2 = - 18/2 = - 9
n₂ = (- b + √Δ)/2a = (- 2 + 16)/2 = 14/2 = 7
Ponieważ n ∈ N⁺ , więc n = 7
a₇ = (7 + 1)² = 8² = 64
b)
bn = n⁴ - 13n² = - 36
n⁴ - 13n² + 36 = 0
Za n² wstawiamy z
z² - 13z + 36 = 0
a = 1 , b = - 13 , c = 36
Δ = b² - 4ac = (- 13)² - 4 * 1 * 36 = 169 - 144 = 25
√Δ = √25 = 5
z₁ = ( - b - √Δ)/2a = (13 - 5)/2 = 8/2 = 4
z₂ = (- b + √Δ)/2a = (13 + 5)/2 = 18/2 = 9
dla z₁
n² = 4
n² - 4 = 0
(n - 2)(n + 2) = 0
n - 2 = 0 ∨ n + 2 = 0
n = 2 ∨ n = - 2
a₂ = 2⁴ - 13 * 2² = 16 - 52 = - 36
Ponieważ n ∈ N⁺ ,więc n = 2
dla z₂
n² = 9
n² - 9 = 0
(n - 3)(n + 3) = 0
n - 3 = 0 ∨ n + 3 = 0
n = 3 ∨ n = - 3
Ponieważ n ∈ N⁺ ,więc n = 3
a₃ = 3⁴ - 13 * 3² = 81 - 117 = - 36
c)
cn = In - 5I + 23
n - 5 + 23 = 28 ∨ n - 5 + 23 = - 28
n + 18 = 28 ∨ n + 18 = - 28
n = 28 - 18 = 10 ∨ n = - 28 - 18 = - 46
Ponieważ n ∈ N⁺ , więc n = 10
c₁₀ = I10 - 5I + 23 = I5I + 23 = 5 + 23 = 28
[tex]a_{n} = (n+1)^{2} = 64\\\\(n+1)^{2} = 64\\\\(n+1)=8^{2} \ \ \ |\sqrt[/tex]
[tex]n+1 = 8 \ \ \ |-1\\\\n = 7\\\\\boxed{a_7 = 64}[/tex]
[tex]b_{n} = n^{4}-13n=-36\\\\n^{4}-13n^{2} =-36\\\\n^{4}-13n^{2}+36 = 0\\\\Niech \ \ t = n^{2}\\\\t^{2}-13t+36 = 0\\\\a = 1, \ b = -13, \ c = 36\\\\\Delta = b^{2}-4ac = (-13)^{2}-4\cdot1\cdot36 = 169-144 = 25\\\\\sqrt{\Delta} = \sqrt{25} = 5\\\\n \in N+\\\\t_1 = \frac{-b-\sqrt{\Delta}}{2a}=\frac{13-5}{2\cdot1}=\frac{8}{2} = 4 \ \ \rightarrow \ \ n = \sqrt{4} =\boxed{ 2}[/tex]
[tex]t_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{13+5}{2} =\frac{18}{2} = 9 \ \ \rightarrow \ \ n=\sqrt{9} =\boxed{ 3}[/tex]
[tex]n \in \{2,3\}\\\\\boxed{b_2 = -36 \ \ oraz \ \ b_3 = -36}[/tex]
[tex]c_{n}=|n-5|+23=28\\\\|n-5| + 23 = 28 \ \ \ |-23\\\\|n-5| = 5\\\\n-5 = 5 \ \vee \ n-5=-5\\\\n\in N+\\\\n = 5+5 \ \vee \ n = -5+5\\\\n = 10 \ \vee \ n = 0 \ \notin \ D\\\\n = 10\\\\\boxed{c_{10} = 28}[/tex]
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Odpowiedź:
a)
an = (n + 1)² ; n ∈ N⁺
(n + 1)² = 64
n² + 2n + 1 = 64
n² + 2n + 1 - 64 = 0
n² + 2n - 63 = 0
a = 1 , b = 2 , c = - 63
Δ = b² - 4ac = 2² - 4 * 1 * (- 63) = 4 + 252 = 256
√Δ = √256 = 16
n₁ = (- b - √Δ)/2a = (- 2 - 16)/2 = - 18/2 = - 9
n₂ = (- b + √Δ)/2a = (- 2 + 16)/2 = 14/2 = 7
Ponieważ n ∈ N⁺ , więc n = 7
a₇ = (7 + 1)² = 8² = 64
b)
bn = n⁴ - 13n² = - 36
n⁴ - 13n² + 36 = 0
Za n² wstawiamy z
z² - 13z + 36 = 0
a = 1 , b = - 13 , c = 36
Δ = b² - 4ac = (- 13)² - 4 * 1 * 36 = 169 - 144 = 25
√Δ = √25 = 5
z₁ = ( - b - √Δ)/2a = (13 - 5)/2 = 8/2 = 4
z₂ = (- b + √Δ)/2a = (13 + 5)/2 = 18/2 = 9
dla z₁
n² = 4
n² - 4 = 0
(n - 2)(n + 2) = 0
n - 2 = 0 ∨ n + 2 = 0
n = 2 ∨ n = - 2
a₂ = 2⁴ - 13 * 2² = 16 - 52 = - 36
Ponieważ n ∈ N⁺ ,więc n = 2
a₂ = 2⁴ - 13 * 2² = 16 - 52 = - 36
dla z₂
n² = 9
n² - 9 = 0
(n - 3)(n + 3) = 0
n - 3 = 0 ∨ n + 3 = 0
n = 3 ∨ n = - 3
Ponieważ n ∈ N⁺ ,więc n = 3
a₃ = 3⁴ - 13 * 3² = 81 - 117 = - 36
c)
cn = In - 5I + 23
n - 5 + 23 = 28 ∨ n - 5 + 23 = - 28
n + 18 = 28 ∨ n + 18 = - 28
n = 28 - 18 = 10 ∨ n = - 28 - 18 = - 46
Ponieważ n ∈ N⁺ , więc n = 10
c₁₀ = I10 - 5I + 23 = I5I + 23 = 5 + 23 = 28
a)
[tex]a_{n} = (n+1)^{2} = 64\\\\(n+1)^{2} = 64\\\\(n+1)=8^{2} \ \ \ |\sqrt[/tex]
[tex]n+1 = 8 \ \ \ |-1\\\\n = 7\\\\\boxed{a_7 = 64}[/tex]
b)
[tex]b_{n} = n^{4}-13n=-36\\\\n^{4}-13n^{2} =-36\\\\n^{4}-13n^{2}+36 = 0\\\\Niech \ \ t = n^{2}\\\\t^{2}-13t+36 = 0\\\\a = 1, \ b = -13, \ c = 36\\\\\Delta = b^{2}-4ac = (-13)^{2}-4\cdot1\cdot36 = 169-144 = 25\\\\\sqrt{\Delta} = \sqrt{25} = 5\\\\n \in N+\\\\t_1 = \frac{-b-\sqrt{\Delta}}{2a}=\frac{13-5}{2\cdot1}=\frac{8}{2} = 4 \ \ \rightarrow \ \ n = \sqrt{4} =\boxed{ 2}[/tex]
[tex]t_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{13+5}{2} =\frac{18}{2} = 9 \ \ \rightarrow \ \ n=\sqrt{9} =\boxed{ 3}[/tex]
[tex]n \in \{2,3\}\\\\\boxed{b_2 = -36 \ \ oraz \ \ b_3 = -36}[/tex]
c)
[tex]c_{n}=|n-5|+23=28\\\\|n-5| + 23 = 28 \ \ \ |-23\\\\|n-5| = 5\\\\n-5 = 5 \ \vee \ n-5=-5\\\\n\in N+\\\\n = 5+5 \ \vee \ n = -5+5\\\\n = 10 \ \vee \ n = 0 \ \notin \ D\\\\n = 10\\\\\boxed{c_{10} = 28}[/tex]