( u n ) ′ = Δ u n = u n + 1 − u n (u_{n})' = \Delta u_{n} = u_{n+1}-u_{n} ( u n ) ′ = Δ u n = u n + 1 − u n ∑ k = 0 n − 1 Δ u k = [ u n ] 0 n \sum_{k=0}^{n-1} \Delta u_{k} = [u_{n}]_{0}^{n} k = 0 ∑ n − 1 Δ u k = [ u n ] 0 n ∑ k = 0 n − 1 k = n ( n − 1 ) 2 \sum_{k=0}^{n-1} k =\frac{n(n-1)}{2} k = 0 ∑ n − 1 k = 2 n ( n − 1 ) ∑ k = 0 n − 1 a k = [ a k ] 0 n a − 1 = a n − 1 a − 1 , a ≠ 0 , 1 \sum_{k=0}^{n-1} a^{k} =\frac{[a^k]^n_0}{a-1} = \frac{a^n-1}{a -1} \quad, \quad a\ne 0,1 k = 0 ∑ n − 1 a k = a − 1 [ a k ] 0 n = a − 1 a n − 1 , a = 0 , 1 n 3 ‾ = n ( n − 1 ) ( n − 2 ) n^{\underline{3}} = n(n-1)(n-2) n 3 = n ( n − 1 ) ( n − 2 ) ( n + 1 ) − 3 ‾ = 1 ( n + 2 ) ( n + 3 ) ( n + 4 ) (n+1)^{\underline{-3}} = \frac{1}{(n+2)(n+3)(n+4)} ( n + 1 ) − 3 = ( n + 2 ) ( n + 3 ) ( n + 4 ) 1 ( a n + b ) r ‾ = Δ ( a n + b ) r + 1 ‾ a ( r + 1 ) (an+b)^{\underline{r}} = \frac{\Delta(an+b)^{\underline{r+1}}}{a(r+1)} ( an + b ) r = a ( r + 1 ) Δ ( an + b ) r + 1 H n = ∑ k = 1 n 1 k , H 0 = 0 H_n = \sum_{k=1}^{n}\frac{1}{k} \quad , H_0 = 0 H n = k = 1 ∑ n k 1 , H 0 = 0 ∑ k = 0 n-1 u k Δ v k = [ u k v k ] 0 n − ∑ k = 0 n − 1 v k + 1 Δ u k \sum_{k=0}^{\textbf{n-1}}u_k\Delta v_k = [u_kv_k]_0^\textbf{n} - \sum_{k=0}^{n-1}v_{k+1}\Delta u_k k = 0 ∑ n-1 u k Δ v k = [ u k v k ] 0 n − k = 0 ∑ n − 1 v k + 1 Δ u k Δ u n v n = ( Δ u n ) v n − u n Δ v n v n v n + 1 \Delta \frac{u_n}{v_n} = \frac{(\Delta u_n)v_n -u_n\Delta v_n}{v_nv_{n+1}} Δ v n u n = v n v n + 1 ( Δ u n ) v n − u n Δ v n u n = u 0 ⋅ n 0 ‾ + 1 1 ! ( Δ u 0 ) n 1 ‾ + 1 2 ! ( Δ 2 u 0 ) n 2 ‾ + ⋯ + 1 k ! ( Δ k u 0 ) n k ‾ u_n = u_0 \cdot n^{\underline 0} + \frac{1}{1!} (\Delta u_0) n^{\underline 1} + \frac{1}{2!} (\Delta^2 u_0) n^{\underline 2} + \dots +
\frac{1}{k!} (\Delta^k u_0) n^{\underline k} u n = u 0 ⋅ n 0 + 1 ! 1 ( Δ u 0 ) n 1 + 2 ! 1 ( Δ 2 u 0 ) n 2 + ⋯ + k ! 1 ( Δ k u 0 ) n k Calcular os números de Stirling de 1ª espécie:
0 1 2 3 k 0 k 1 k 2 k 3 0 k 0 ‾ 1 0 0 0 c k 1 ‾ a b 0 0 2 k 2 ‾ 0 d 1 0 3 k 3 ‾ 0 2 − 3 1 \begin{array}{ c c c c c c }
& & 0 & 1 & 2 & 3\\
& & k^{0} & k^{1} & k^{2} & k^{3}\\
0 & k^{\underline{0}} & 1 & 0 & 0 & 0\\
\smartcolor{red}{\mathbf{c}} & k^{\underline{1}} & \smartcolor{red}{\mathbf{a}} & \smartcolor{red}{\mathbf{b}} & 0 & 0\\
2 & k^{\underline{2}} & 0 & \smartcolor{red}{\mathbf{d}} & 1 & 0\\
3 & k^{\underline{3}} & 0 & 2 & -3 & 1
\end{array} 0 c 2 3 k 0 k 1 k 2 k 3 0 k 0 1 a 0 0 1 k 1 0 b d 2 2 k 2 0 0 1 − 3 3 k 3 0 0 0 1 0 1 2 3 4 5 k 0 k 1 k 2 k 3 k 4 k 5 0 k 0 ‾ 1 0 0 0 0 0 1 k 1 ‾ 0 1 0 0 0 0 2 k 2 ‾ 0 − 1 1 0 0 0 3 k 3 ‾ 0 2 − 3 1 0 0 4 k 4 ‾ 0 − 6 11 − 6 1 0 5 k 5 ‾ 0 24 − 50 35 − 10 1 \begin{array}{ c c c c c c c c }
& & 0 & 1 & 2 & 3 & 4 & 5\\
& & k^{0} & k^{1} & k^{2} & k^{3} & k^{4} & k^{5}\\
0 & k^{\underline{0}} & 1 & 0 & 0 & 0 & 0 & 0\\
1 & k^{\underline{1}} & 0 & 1 & 0 & 0 & 0 & 0\\
2 & k^{\underline{2}} & 0 & -1 & 1 & 0 & 0 & 0\\
3 & k^{\underline{3}} & 0 & 2 & -3 & 1 & 0 & 0\\
4 & k^{\underline{4}} & 0 & -6 & 11 & -6 & 1 & 0\\
5 & k^{\underline{5}} & 0 & 24 & -50 & 35 & -10 & 1
\end{array} 0 1 2 3 4 5 k 0 k 1 k 2 k 3 k 4 k 5 0 k 0 1 0 0 0 0 0 1 k 1 0 1 − 1 2 − 6 24 2 k 2 0 0 1 − 3 11 − 50 3 k 3 0 0 0 1 − 6 35 4 k 4 0 0 0 0 1 − 10 5 k 5 0 0 0 0 0 1 Calcular os números de Stirling de 2ª espécie:
0 1 b 3 k 0 ‾ k 1 ‾ k 2 ‾ k 3 ‾ 0 k 0 1 0 0 0 1 k 1 0 1 0 0 2 k 2 0 c a 0 3 k 3 0 1 d 1 \begin{array}{ c c c c c c }
& & 0 & 1 & \smartcolor{red}{\mathbf{b}} & 3\\
& & k^{\underline{0}} & k^{\underline{1}} & k^{\underline{2}} & k^{\underline{3}}\\
0 & k^{0} & 1 & 0 & 0 & 0\\
1 & k^{1} & 0 & 1 & 0 & 0\\
2 & k^{2} & 0 & \smartcolor{red}{\mathbf{c}} & \smartcolor{red}{\mathbf{a}} & 0\\
3 & k^{3} & 0 & 1 & \smartcolor{red}{\mathbf{d}} & 1
\end{array} 0 1 2 3 k 0 k 1 k 2 k 3 0 k 0 1 0 0 0 1 k 1 0 1 c 1 b k 2 0 0 a d 3 k 3 0 0 0 1 0 1 2 3 4 5 k 0 ‾ k 1 ‾ k 2 ‾ k 3 ‾ k 4 ‾ k 5 ‾ 0 k 0 1 0 0 0 0 0 1 k 1 0 1 0 0 0 0 2 k 2 0 1 1 0 0 0 3 k 3 0 1 3 1 0 0 4 k 4 0 1 7 6 1 0 5 k 5 0 1 15 25 10 1 \begin{array}{ c c c c c c c c }
& & 0 & 1 & 2 & 3 & 4 & 5\\
& & k^{\underline{0}} & k^{\underline{1}} & k^{\underline{2}} & k^{\underline{3}} & k^{\underline{4}} & k^{\underline{5}}\\
0 & k^{0} & 1 & 0 & 0 & 0 & 0 & 0\\
1 & k^{1} & 0 & 1 & 0 & 0 & 0 & 0\\
2 & k^{2} & 0 & 1 & 1 & 0 & 0 & 0\\
3 & k^{3} & 0 & 1 & 3 & 1 & 0 & 0\\
4 & k^{4} & 0 & 1 & 7 & 6 & 1 & 0\\
5 & k^{5} & 0 & 1 & 15 & 25 & 10 & 1
\end{array} 0 1 2 3 4 5 k 0 k 1 k 2 k 3 k 4 k 5 0 k 0 1 0 0 0 0 0 1 k 1 0 1 1 1 1 1 2 k 2 0 0 1 3 7 15 3 k 3 0 0 0 1 6 25 4 k 4 0 0 0 0 1 10 5 k 5 0 0 0 0 0 1 S n + 1 = ∑ k = 0 n + 1 2 k = ∑ k = 0 n 2 k + 2 n + 1 = S n + 2 n + 1 S_{n+1} = \sum^{n+1}_{k=0} 2^k = \sum^n_{k=0} 2^k + 2^{n+1} = S_n + 2^{n+1} S n + 1 = k = 0 ∑ n + 1 2 k = k = 0 ∑ n 2 k + 2 n + 1 = S n + 2 n + 1 S n + 1 = ∑ k = 0 n + 1 2 k = 2 0 + ∑ k = 1 n + 1 2 k = 1 + ∑ k = 0 n 2 k + 1 = 1 + ∑ k = 0 n ( 2 k × 2 ) = = 1 + 2 ∑ k = 0 n 2 k = 1 + 2 S n S_{n+1} =\sum ^{n+1}_{k=0} 2^{k} =2^{0} +\sum ^{n+1}_{k=1} 2^{k} =1+\sum ^{n}_{k=0} 2^{k+1} =1+\sum ^{n}_{k=0}\left( 2^{k} \times 2\right) =\\
=1+2\sum ^{n}_{k=0} 2^{k} =1+2S_{n} S n + 1 = k = 0 ∑ n + 1 2 k = 2 0 + k = 1 ∑ n + 1 2 k = 1 + k = 0 ∑ n 2 k + 1 = 1 + k = 0 ∑ n ( 2 k × 2 ) = = 1 + 2 k = 0 ∑ n 2 k = 1 + 2 S n Basta agora igualar estes dois resultados e isolar S n S_n S n
Fazer a Perturbação Direta mas dentro do Somatório multiplicar a expressão por k .
Para calcularmos o valor de ∑ k = 0 n k \displaystyle \sum^n_{k=0}k k = 0 ∑ n k
calculamos o valor de ∑ k = 0 n k 2 \displaystyle \sum^n_{k=0}k^2 k = 0 ∑ n k 2
Que P ( 0 ) P(0) P ( 0 )
∀ n ∈ N ( P ( n ) e ˊ verdadeiro ⟹ P ( n + 1 ) e ˊ verdadeiro ) \forall_{n \in \mathbb N} \ (P(n) \text{ é verdadeiro}\implies P(n+1) \text{ é verdadeiro }) ∀ n ∈ N ( P ( n ) e ˊ verdadeiro ⟹ P ( n + 1 ) e ˊ verdadeiro )
Definição de indução simples:
Se P ( m ) , P ( m + 1 ) , P ( m + 2 ) , … ( m ∈ N ) P(m), P(m+1), P(m+2),\dots ( m \in \mathbb N) P ( m ) , P ( m + 1 ) , P ( m + 2 ) , … ( m ∈ N )
P ( m ) P(m) P ( m ) ∀ n ∈ N \forall_{n \in \mathbb N} ∀ n ∈ N ( P ( n ) (P(n) ( P ( n ) ⟶ P ( n + 1 ) \longrightarrow P(n+1) ⟶ P ( n + 1 ) ) ) )
Definição de indução completa:
Se P ( m ) , P ( m + 1 ) , … P(m), P(m+1), \dots P ( m ) , P ( m + 1 ) , …
P ( m ) , P ( m + 1 ) , … P(m), P(m+1), \dots P ( m ) , P ( m + 1 ) , … ∀ n ∈ N , P ( m ) , … , P ( m + γ ) , P ( m + γ + 1 ) , … , P ( m + γ + n ) ⟶ P ( m + γ + n + 1 ) \forall_{n \in \mathbb N}, P(m), \dots, P(m+ \gamma), P(m+\gamma+1), \dots, P(m+\gamma+n) \longrightarrow P(m+\gamma+n+1) ∀ n ∈ N , P ( m ) , … , P ( m + γ ) , P ( m + γ + 1 ) , … , P ( m + γ + n ) ⟶ P ( m + γ + n + 1 )
então ∀ n ∈ N P ( n ) \forall_{n \in \mathbb N} P(n) ∀ n ∈ N P ( n )
Retirado da série 1, 1.2.1 a.
∑ k = 0 n 4 k \sum^n_{k=0} 4^k k = 0 ∑ n 4 k Resolução Neste caso, podemos fazer a diferença finita Δ 4 k \Delta 4^k Δ 4 k 4 k 4^k 4 k
Δ 4 k = 4 k + 1 − 4 k = 4 k ( 4 − 1 ) = 3 × 4 k \Delta 4^k = 4^{k+1} - 4^k = 4^k(4-1) = 3 \times 4^k Δ 4 k = 4 k + 1 − 4 k = 4 k ( 4 − 1 ) = 3 × 4 k 4 k = Δ 4 k 3 4^k = \Delta \frac{4^k}{3} 4 k = Δ 3 4 k Então:
∑ k = 0 n 4 k = ∑ k = 0 n Δ 4 k 3 = 1 3 [ 4 k ] 0 n + 1 = 1 3 ( 4 n + 1 − 4 0 ) = 1 3 ( 4 n + 1 − 1 ) \begin{aligned}
\sum^n_{k=0} 4^k &= \sum^n_{k=0} \Delta \frac{4^k}{3}\\
&= \frac{1}{3} \left[ 4^k \right]^{n+1}_0\\
&= \frac{1}{3}(4^{n+1} - 4^0)\\
&= \frac{1}{3}(4^{n+1} - 1)
\end{aligned} k = 0 ∑ n 4 k = k = 0 ∑ n Δ 3 4 k = 3 1 [ 4 k ] 0 n + 1 = 3 1 ( 4 n + 1 − 4 0 ) = 3 1 ( 4 n + 1 − 1 ) Retirado da série 1, 1.2.1 d.
∑ k = 0 n k × 4 k \sum^n_{k=0} k\times 4^k k = 0 ∑ n k × 4 k Resolução Neste caso, podemos tomar u k = k u_k=k u k = k Δ v k \Delta v_k Δ v k Δ 4 k \Delta 4^k Δ 4 k 4 k 4^k 4 k Δ 4 k \Delta 4^k Δ 4 k
Δ 4 k = 4 k + 1 − 4 k = 4 k ( 4 − 1 ) = 3 × 4 k \Delta 4^k = 4^{k+1} - 4^k = 4^k(4-1) = 3 \times 4^k Δ 4 k = 4 k + 1 − 4 k = 4 k ( 4 − 1 ) = 3 × 4 k 4 k = Δ 4 k 3 4^k = \Delta \frac{4^k}{3} 4 k = Δ 3 4 k Então:
∑ k = 0 n k × 4 k = ∑ k = 0 n k × Δ 4 k 3 = 1 3 ( [ k × 4 k ] 0 n + 1 − ∑ k = 0 n 4 k + 1 ⋅ Δ k ) = 1 3 ( ( n + 1 ) 4 n + 1 − 0 − 4 ∑ k = 0 n 4 k × 1 ) = 1 3 ( ( n + 1 ) 4 n + 1 − 4 ( 1 3 ( 4 n + 1 − 1 ) ) ) \begin{aligned}
\sum^n_{k=0} k\times 4^k &= \sum^n_{k=0} k \times \Delta \frac{4^k}{3}\\
&= \frac{1}{3} \left(\left[k\times 4^k \right]^{n+1}_0 - \sum^n_{k=0} 4^{k+1} \cdot \Delta k \right)\\
&= \frac{1}{3} \left((n+1)4^{n+1} - 0 - 4\sum^n_{k=0} 4^{k} \times 1 \right)\\
&= \frac{1}{3} \left((n+1)4^{n+1} - 4\left(\frac{1}{3}\left(4^{n+1}-1\right)\right) \right)\\
\end{aligned} k = 0 ∑ n k × 4 k = k = 0 ∑ n k × Δ 3 4 k = 3 1 ( [ k × 4 k ] 0 n + 1 − k = 0 ∑ n 4 k + 1 ⋅ Δ k ) = 3 1 ( ( n + 1 ) 4 n + 1 − 0 − 4 k = 0 ∑ n 4 k × 1 ) = 3 1 ( ( n + 1 ) 4 n + 1 − 4 ( 3 1 ( 4 n + 1 − 1 ) ) ) Este último passo foi feito através do exemplo anterior.
Retirado da série 1, 1.2.3 g.
Usa-se este método quando não se consegue imediatamente exprimir o quociente numa potência fatorial.
∑ k = 2 n 1 ( k + 2 ) ( k + 5 ) \sum^{n}_{k=2} \frac{1}{(k+2)(k+5)} k = 2 ∑ n ( k + 2 ) ( k + 5 ) 1 Resolução Ver a resolução do 1.2.3 g., que está na página 2 deste PDF .
Retirado da série 1, 1.2.5 a.
∑ k = 0 n − 1 k 2 k ( k + 1 ) ( k + 2 ) \sum^{n-1}_{k=0} \frac{k2^k}{(k+1)(k+2)} k = 0 ∑ n − 1 ( k + 1 ) ( k + 2 ) k 2 k Resolução Começar por tentar "adivinhar" a expressão que ao derivar dá aquilo que está "dentro" do somatório.Em alguns casos, podemos usar variáveis que depois alteramos por um valor, para termos uma maior probabilidade de acertar.
Δ 2 k k + 1 = ( Δ 2 k ) ( k + 1 ) − 2 k ( Δ k ) ( k + 1 ) ( k + 2 ) = 2 k k + 2 k − 2 k ( k + 1 ) ( k + 2 ) = k 2 k ( k + 1 ) ( k + 2 ) \Delta \frac{2^k}{k+1} = \frac{(\Delta 2^k)(k+1)-2^k(\Delta k)}{(k+1)(k+2)}=\frac{2^k k + 2^k - 2^k}{(k+1)(k+2)} = \frac{k 2^k}{(k+1)(k+2)} Δ k + 1 2 k = ( k + 1 ) ( k + 2 ) ( Δ 2 k ) ( k + 1 ) − 2 k ( Δ k ) = ( k + 1 ) ( k + 2 ) 2 k k + 2 k − 2 k = ( k + 1 ) ( k + 2 ) k 2 k ∑ k = 0 n − 1 k 2 k ( k + 1 ) ( k + 2 ) = [ 2 k k + 1 ] 0 n = 2 n n + 1 − 2 0 0 + 1 = 2 n n + 1 − 1 \sum^{n-1}_{k=0} \frac{k2^k}{(k+1)(k+2)} = \left[\frac{2^k}{k+1} \right]^n_0 = \frac{2^n}{n+1}- \frac{2^0}{0+1} = \frac{2^n}{n+1}-1 k = 0 ∑ n − 1 ( k + 1 ) ( k + 2 ) k 2 k = [ k + 1 2 k ] 0 n = n + 1 2 n − 0 + 1 2 0 = n + 1 2 n − 1
