Suites explicites et définies par récurrence : calculs de termes

\( u_0 = 3 \times 0^2 - 0 + 5 = \) \( 5 \)

\( u_1 = 3 \times 1^2 - 1 + 5 = 3 - 1 + 5 = \) \( 7 \)

\( u_5 = 3 \times 5^2 - 5 + 5 = 3 \times 25 - 5 + 5 = \) \( 75 \)

\( v_0 = \dfrac{4 \times 0 - 1}{0 + 3} = \) \( -\dfrac{1}{3} \)

\( v_1 = \dfrac{4 \times 1 - 1}{1 + 3} = \) \( \dfrac{3}{4} \)

\( v_7 = \dfrac{4 \times 7 - 1}{7 + 3} = \dfrac{28 - 1}{10} = \) \( \dfrac{27}{10} \)

\( w_0 = 5 \times 0 + (-1)^{0+1} = 0 + (-1)^1 = \) \( -1 \)

\( w_1 = 5 \times 1 + (-1)^{1+1} = 5 + (-1)^2 = 5 + 1 = \) \( 6 \)

\( w_2 = 5 \times 2 + (-1)^{2+1} = 10 + (-1)^3 = 10 - 1 = \) \( 9 \)

\( u_0 = \) \( 10 \)

\( u_1 = -2 \times u_0 + 3 = -2 \times 10 + 3 = \) \( -17 \)

\( u_2 = -2 \times u_1 + 3 = -2 \times (-17) + 3 = 34 + 3 = \) \( 37 \)

\( v_0 = \) \( 1 \)

\( v_1 = 2 \times v_0 - 0 + 1 = 2 \times 1 + 1 = \) \( 3 \)

\( v_2 = 2 \times v_1 - 1 + 1 = 2 \times 3 = \) \( 6 \)

\( w_0 = \) \( 2 \)

\( w_1 = \dfrac{w_0}{w_0 + 1} = \dfrac{2}{2+1} = \) \( \dfrac{2}{3} \)

\( w_2 = \dfrac{w_1}{w_1 + 1} = \dfrac{2/3}{2/3 + 1} = \dfrac{2/3}{5/3} = \) \( \dfrac{2}{5} \)

\( v_{n-1} = \dfrac{3 \times (n-1) - 1}{(n-1)^2 + 1} = \dfrac{3n - 3 - 1}{n^2 - 2n + 1 + 1} = \) \( \dfrac{3n - 4}{n^2 - 2n + 2} \)

\( v_{n+2} = \dfrac{3 \times (n+2) - 1}{(n+2)^2 + 1} = \dfrac{3n + 6 - 1}{n^2 + 4n + 4 + 1} = \) \( \dfrac{3n + 5}{n^2 + 4n + 5} \)

\( v_{2n} = \dfrac{3 \times (2n) - 1}{(2n)^2 + 1} = \) \( \dfrac{6n - 1}{4n^2 + 1} \)

\( w_{n-1} = 2 \times (n-1)^2 - 3 \times (n-1) + 1 = 2 \times (n^2 - 2n + 1) - 3n + 3 + 1 = 2n^2 - 4n + 2 - 3n + 4 = \) \( 2n^2 - 7n + 6 \)

\( w_{n+1} = 2 \times (n+1)^2 - 3 \times (n+1) + 1 = 2 \times (n^2 + 2n + 1) - 3n - 3 + 1 = 2n^2 + 4n + 2 - 3n - 2 = \) \( 2n^2 + n \)

\( w_{2n} = 2 \times (2n)^2 - 3 \times (2n) + 1 = 2 \times (4n^2) - 6n + 1 = \) \( 8n^2 - 6n + 1 \)

\( w_{2n+1} = 2 \times (2n+1)^2 - 3 \times (2n+1) + 1 = 2 \times (4n^2 + 4n + 1) - 6n - 3 + 1 = 8n^2 + 8n + 2 - 6n - 2 = \) \( 8n^2 + 2n \)

La formule de récurrence \( u_{k+1} = 3u_k - 2 \) est vraie pour tout \( k \in \mathbb{N} \).

On l'applique en remplaçant \( k \) par \( (n+1) \) :

\( u_{n+2} = 3 \times u_{n+1} - 2 \)

On part du résultat précédent : \( u_{n+2} = 3 \times u_{n+1} - 2 \).

On remplace \( u_{n+1} \) par son expression en fonction de \( u_n \) (qui est \( 3u_n - 2 \) :

\( u_{n+2} = 3 \times (3u_n - 2) - 2 = 9u_n - 6 - 2 = \) \( 9u_n - 8 \)

La formule de récurrence \( v_{k+1} = (k+1)v_k + 4 \) est vraie pour tout \( k \in \mathbb{N} \).

On l'applique en remplaçant \( k \) par \( (n-1) \) (valable pour \( n-1 \ge 0 \), donc \( n \ge 1 \)) :

\( v_n = ( (n-1)+1 ) \times v_{n-1} + 4 = n \times v_{n-1} + 4 \)

On part de la formule \( v_{k+1} = (k+1)v_k + 4 \) et on l'applique pour \( k=n+1 \) :

\( v_{n+2} = ((n+1)+1) \times v_{n+1} + 4 = (n+2) \times v_{n+1} + 4 \)

On remplace \( v_{n+1} \) par son expression en fonction de \( v_n \) (qui est \( (n+1)v_n + 4 \)) :

\( v_{n+2} = (n+2) \times [ (n+1) \times v_n + 4 ] + 4 \)

\( v_{n+2} = (n+2) \times (n+1) \times v_n + 4 \times (n+2) + 4 \)

\( v_{n+2} = (n+2) \times (n+1) \times v_n + 4n + 8 + 4 = \) \( (n^2+3n+2) \times v_n + 4n + 12 \)