∑ k = 1 n − 1 ( k − 2 ) ( k − 1 ) ( k + 1 ) ( k + 2 ) ( k + 3 ) \sum_{k=1}^{n-1}\frac{(k-2)(k-1)}{(k+1)(k+2)(k+3)} k = 1 ∑ n − 1 ( k + 1 ) ( k + 2 ) ( k + 3 ) ( k − 2 ) ( k − 1 ) De início não parece ser possível haver uma resolução direta devido ao que se tem no numerador. Em exercícios deste tipo usa-se o que se chama o método das pontas , que consiste em eliminar os fatores mais exteriores do denominador. Por exemplo:
( k − 2 ) ( k − 1 ) = k 2 − 3 k + 2 (k-2)(k-1)=k^2-3k+2 ( k − 2 ) ( k − 1 ) = k 2 − 3 k + 2 Queremos transformar isto nalgo que tenha fatores, por exemplo, ( k + 3 ) (k+3) ( k + 3 ) k + 1 k+1 k + 1 ( k + 3 ) ( k + 2 ) (k+3)(k+2) ( k + 3 ) ( k + 2 )
1 − 3 2 − 3 − 3 18 1 − 6 20 − 2 − 2 1 − 8 \begin{array}{ c | c c c c c c }
& \\
& 1 & -3 & 2 \\
&\\
\ -3 & & -3 & 18 \\
\hline &\\
& 1 & -6 & \smartcolor{orange}{20} \\
&\\
-2 & & -2 & \\
\hline &\\
& 1 & \smartcolor{orange}{-8} \\
\end{array} − 3 − 2 1 1 1 − 3 − 3 − 6 − 2 − 8 2 18 20 O laranja é o Resto da Divisão.
⟶ 1 ⋅ ( k + 3 ) ( k + 2 ) − 8 ( k + 3 ) + 20 \longrightarrow 1\cdot(k+3)(k+2)\smartcolor{orange}{-8}(k+3)\smartcolor{orange}{+20} ⟶ 1 ⋅ ( k + 3 ) ( k + 2 ) − 8 ( k + 3 ) + 20 e o somatório original vem como:
∑ k = 1 n − 1 ( k + 3 ) ( k + 2 ) ( k + 1 ) ( k + 2 ) ( k + 3 ) − 8 ∑ k = 1 n − 1 ( k + 3 ) ( k + 1 ) ( k + 2 ) ( k + 3 ) + 20 ∑ k = 1 n − 1 1 ( k + 1 ) ( k + 2 ) ( k + 3 ) = ∑ k = 1 n − 1 Δ H k + [ 8 k + 1 − 10 ( k + 1 ) ( k + 2 ) ] 1 n = H n − 1 + 8 n + 1 − 4 − 10 ( n + 1 ) ( n + 2 ) + 5 3 \sum_{k=1}^{n-1}\frac{(k+3)(k+2)}{(k+1)(k+2)(k+3)} -\\ 8\sum_{k=1}^{n-1}\frac{(k+3)}{(k+1)(k+2)(k+3)}+\\20\sum_{k=1}^{n-1}\frac{1}{(k+1)(k+2)(k+3)} =\\
\sum_{k=1}^{n-1}\Delta H_k + \left[\frac{8}{k+1}-\frac{10}{(k+1)(k+2)}\right]_1^n = \\
H_n-1+\frac{8}{n+1}-4-\frac{10}{(n+1)(n+2)} + \frac 5 3 k = 1 ∑ n − 1 ( k + 1 ) ( k + 2 ) ( k + 3 ) ( k + 3 ) ( k + 2 ) − 8 k = 1 ∑ n − 1 ( k + 1 ) ( k + 2 ) ( k + 3 ) ( k + 3 ) + 20 k = 1 ∑ n − 1 ( k + 1 ) ( k + 2 ) ( k + 3 ) 1 = k = 1 ∑ n − 1 Δ H k + [ k + 1 8 − ( k + 1 ) ( k + 2 ) 10 ] 1 n = H n − 1 + n + 1 8 − 4 − ( n + 1 ) ( n + 2 ) 10 + 3 5 Pretende-se estudar a diferença finita de expressões do tipo u n v n u_n v_n u n v n
tem-se:
Δ ( u n v n ) = u n + 1 v n + 1 − u n v n = u n + 1 v n + 1 + u n v n + 1 − u n v n + 1 ‾ − u n v n = v n + 1 ( u n + 1 − u n ) + u n ( v n + 1 − v n ) = v n + 1 Δ u n + u n Δ v n ⟶ u n Δ v n = Δ ( u n v n ) − v n + 1 Δ u n \Delta(u_nv_n)=u_{n+1}v_{n+1}-u_nv_n =\\
u_{n+1}v_{n+1} + \underline{u_nv_{n+1} -u_nv_{n+1}} - u_nv_n = \\
v_{n+1}(u_{n+1}-u_n)+u_n(v_{n+1}-v_n) = \\
v_{n+1}\Delta u_n + u_n\Delta v_n \longrightarrow \\
\\
\huge{u_n\Delta v_n = \Delta(u_nv_n) - v_{n+1}\Delta u_n} Δ ( u n v n ) = u n + 1 v n + 1 − u n v n = u n + 1 v n + 1 + u n v n + 1 − u n v n + 1 − u n v n = v n + 1 ( u n + 1 − u n ) + u n ( v n + 1 − v n ) = v n + 1 Δ u n + u n Δ v n ⟶ u n Δ v n = Δ ( u n v n ) − v n + 1 Δ u n Pelo que se pode aplicar facilmente:
∑ 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 Exemplo:
∑ k = 0 n − 1 k H k = ∑ k = 0 n − 1 H k Δ k 2 ‾ 2 = [ k 2 ‾ 2 H k ] 0 n − ∑ k = 0 n − 1 ( k ( k + 1 ) 2 ⋅ 1 k + 1 ) \sum_{k=0}^{n-1}kH_k = \sum_{k=0}^{n-1}H_k\Delta\frac{k^{\underline{2}}}{2} = \left[\frac{k^{\underline{2}}}{2}H_k\right]_0^n - \sum_{k=0}^{n-1}\left(\frac{k(k+1)}2\cdot\frac{1}{k+1}\right) k = 0 ∑ n − 1 k H k = k = 0 ∑ n − 1 H k Δ 2 k 2 = [ 2 k 2 H k ] 0 n − k = 0 ∑ n − 1 ( 2 k ( k + 1 ) ⋅ k + 1 1 ) Δ u n v n = u n + 1 v n + 1 − u n v n = u n + 1 v n + u n v n − u n v n ‾ − u n v n + 1 v n v n + 1 = ( Δ u n ) v n − u n Δ v n v n v n + 1 \Delta \frac{u_n}{v_n} = \frac{u_{n+1}}{v_{n+1}} - \frac{u_n}{v_n} = \\
\frac{u_{n+1}v_n+\underline{u_nv_n-u_nv_n} -u_nv_{n+1}}{v_nv_{n+1}} = \\
\huge\frac{(\Delta u_n)v_n -u_n\Delta v_n}{v_nv_{n+1}} Δ v n u n = v n + 1 u n + 1 − v n u n = v n v n + 1 u n + 1 v n + u n v n − u n v n − u n v n + 1 = v n v n + 1 ( Δ u n ) v n − u n Δ v n Exemplos
∑ k = 0 n − 1 k 2 k ( k + 1 ) ( k + 2 ) = ? \sum_{k=0}^{n-1}\frac{k2^k}{(k+1)(k+2)} = \quad ? k = 0 ∑ n − 1 ( k + 1 ) ( k + 2 ) k 2 k = ? Repare-se:
Δ 2 k k + 1 = 2 k ( k + 1 ) − 2 k ( 1 ) ( k + 1 ) ( k + 2 ) = k 2 k ( k + 1 ) ( k + 2 ) \Delta\frac{2^k}{k+1} = \frac{2^k(k+1)-2^k(1)}{(k+1)(k+2)}=\frac{k2^k}{(k+1)(k+2)} Δ k + 1 2 k = ( k + 1 ) ( k + 2 ) 2 k ( k + 1 ) − 2 k ( 1 ) = ( k + 1 ) ( k + 2 ) k 2 k logo:
∑ k = 0 n − 1 k 2 k ( k + 1 ) ( k + 2 ) = ∑ k = 0 n − 1 Δ 2 k k + 1 = [ 2 k k + 1 ] 0 n = 2 n n + 1 − 1 \sum_{k=0}^{n-1}\frac{k2^k}{(k+1)(k+2)} = \sum_{k=0}^{n-1}\Delta\frac{2^k}{k+1} = \left[\frac{2^k}{k+1}\right]_0^n =\frac{2^{n}}{n+1} -1 k = 0 ∑ n − 1 ( k + 1 ) ( k + 2 ) k 2 k = k = 0 ∑ n − 1 Δ k + 1 2 k = [ k + 1 2 k ] 0 n = n + 1 2 n − 1
∑ k = 0 n − 1 k 2 k ( k + 2 ) ! = ? \sum_{k=0}^{n-1}\frac{k2^k}{(k+2)!} = \quad ? k = 0 ∑ n − 1 ( k + 2 )! k 2 k = ? Repara-se:
Δ 2 k ( k + 1 ) ! = 2 k ( k + 1 ) ! − 2 k ( k + 1 ) ( k + 1 ) ! ( k + 1 ) ! ( k + 2 ) ! = = 2 k − 2 k ( k + 1 ) ( k + 2 ) ! = 2 k ( 1 − k − 1 ) ( k + 2 ) ! = − k 2 k ( k + 2 ) ! \Delta \frac{2^{k}}{(k+1)!} =\frac{2^{k} (k+1)!-2^{k} (k+1)(k+1)!}{(k+1)!(k+2)!} =\\
=\frac{2^{k} -2^{k} (k+1)}{(k+2)!} =\frac{2^{k} (1-k-1)}{(k+2)!} =-\frac{k2^{k}}{(k+2)!} Δ ( k + 1 )! 2 k = ( k + 1 )! ( k + 2 )! 2 k ( k + 1 )! − 2 k ( k + 1 ) ( k + 1 )! = = ( k + 2 )! 2 k − 2 k ( k + 1 ) = ( k + 2 )! 2 k ( 1 − k − 1 ) = − ( k + 2 )! k 2 k
∑ k = 0 n − 1 2 k − 1 2 k − 1 = ? \sum_{k=0}^{n-1}\frac{2k-1}{2^{k-1}} = \quad ? k = 0 ∑ n − 1 2 k − 1 2 k − 1 = ? Repara-se:
Δ a k + b 2 k − 1 = a 2 k − 1 − ( a k + b ) 2 k − 1 2 k − 1 2 k = a − ( a k + b ) 2 k = − a 2 k + a − b 2 2 k − 1 \Delta\frac{ak+b}{2^{k-1}} = \frac{a2^{k-1}-(ak+b)2^{k-1}}{2^{k-1}2^{k}}= \frac{a-(ak+b)}{2^{k}} = \frac{-\frac a 2k+\frac{a-b} 2}{2^{k-1}} Δ 2 k − 1 ak + b = 2 k − 1 2 k a 2 k − 1 − ( ak + b ) 2 k − 1 = 2 k a − ( ak + b ) = 2 k − 1 − 2 a k + 2 a − b { − a 2 = 2 a − b 2 = − 1 ⇔ { a = 4 b = − 2 \begin{dcases}-\frac a 2 = 2\\\frac{a-b} 2 = -1\end{dcases} \Leftrightarrow \begin{dcases}a=4\\b = -2\end{dcases} ⎩ ⎨ ⎧ − 2 a = 2 2 a − b = − 1 ⇔ { a = 4 b = − 2 logo
Δ − 4 k − 2 2 k − 1 = 2 k − 1 2 k − 1 \Delta \frac{-4k-2}{2^{k-1}} = \frac{2k-1}{2^{k-1}} Δ 2 k − 1 − 4 k − 2 = 2 k − 1 2 k − 1