Řešení úloh 12–14 z cvičení DIM2

12g

x_{n + 2} + 6 x_{n + 1} + 9 x_{n} = 36 (- 3)^{n}

λ^{2} + 6 λ + 9 = 0

{(λ + 3)}^{2} = 0

h_{n} = α {(- 3)}^{n} + β n {(- 3)}^{n}

S t u p e \overset{ˇ}{n} p o l y n o m u = 0

r = ν (- 3) = 2

p_{n} = γ n^{2} {(- 3)}^{n}

γ {(n + 2)}^{2} {(- 3)}^{n + 2} + 6 γ {(n + 1)}^{2} {(- 3)}^{n + 1} + 9 γ n^{2} {(- 3)}^{n} = 36 {(- 3)}^{n}

γ (9 {(n + 2)}^{2} - 18 {(n + 1)}^{2} + 9 n^{2}) = 36

γ = 2

x_{n} = α {(- 3)}^{n} + β n {(- 3)}^{n} + 2 n^{2} {(- 3)}^{n} = (α + β n + 2 n^{2}) {(- 3)}^{n}

x_{0} = α = 0

x_{1} = - 3 (α + β + 2) = 3 ⟹ β = - 3

x_{n} = n (2 n - 3) {(- 3)}^{n}

y_{n + 2} + 8 y_{n + 1} + 12 y_{n} = e^{n}

λ^{2} + 8 λ + 12 = 0

(λ + 2) (λ + 6) = 0

h_{n} = α {(- 2)}^{n} + β {(- 6)}^{n}

S t u p e \overset{ˇ}{n} p o l y n o m u = 0

r = ν (e) = 0

p_{n} = γ e^{n}

γ e^{n + 2} + 8 γ e^{n + 1} + 12 γ e^{n} = e^{n}

γ (e^{2} + 8 e + 12) = 1

γ = \frac{1}{e^{2} + 8 e + 12}

x_{n} = α {(- 2)}^{n} + β {(- 6)}^{n} + \frac{e^{n}}{e^{2} + 8 e + 12}; α, β \in ℂ

s_{n + 1} - s_{n} = \frac{{(n + 1)}^{2}}{2^{n + 1}}

s_{n + 1} - s_{n} = (\frac{1}{2} n^{2} + n + \frac{1}{2}) {(\frac{1}{2})}^{n}

λ - 1 = 0

h_{n} = α

S t u p e \overset{ˇ}{n} p o l y n o m u = 2

r = ν (\frac{1}{2}) = 0

p_{n} = (β n^{2} + γ n + δ) {(\frac{1}{2})}^{n}

(β {(n + 1)}^{2} + γ (n + 1) + δ) {(\frac{1}{2})}^{n + 1} - (β n^{2} + γ n + δ) {(\frac{1}{2})}^{n} = (\frac{1}{2} n^{2} + n + \frac{1}{2}) {(\frac{1}{2})}^{n}

(β {(n + 1)}^{2} + γ (n + 1) + δ) - (2 β n^{2} + 2 γ n + 2 δ) = (n^{2} + 2 n + 1)

Koeficienty u $n^{2}$

β - 2 β = 1 ⟹ β = - 1

Koeficienty u $n^{1}$

2 β + γ - 2 γ = 2 ⟹ γ = - 4

Koeficienty u $n^{0}$

β + γ + δ - 2 δ = 1 ⟹ δ = - 6

s_{n} = α - (n^{2} + 4 n + 6) {(\frac{1}{2})}^{n}

Počáteční podmínka: $s_{0} = 0$

s_{0} = α - 6 = 0 ⟹ α = 6

\sum_{k = 1}^{n} \frac{k^{2}}{2^{k}} = 6 - (n^{2} + 4 n + 6) {(\frac{1}{2})}^{n}