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Soit ( u n ) n ∈ N (u_{n})_{n \in \mathbb{N}} ( u n ) n ∈ N ∣ u 0 = 0 , u 1 = 1 , ∀ n ∈ N , u n + 2 = 5 u n + 1 − 6 u n . \left|
\begin{array}{l}
u_0 = 0,\\
u_1 = 1,\\
\forall n\in\mathbb{N},\ u_{n+2} = 5u_{n+1} - 6u_n.
\end{array}
\right. u 0 = 0 , u 1 = 1 , ∀ n ∈ N , u n + 2 = 5 u n + 1 − 6 u n .
Montrez par récurrence double que ∀ n ∈ N , u n = 3 n − 2 n . \forall n \in \mathbb{N}, \, u_{n} = 3^{n} - 2^{n}. ∀ n ∈ N , u n = 3 n − 2 n .
Solutions 1 Reveal solutions Are you sure? Give it a try first. Pour n ∈ N n\in \mathbb{N} n ∈ N P ( n ) = P(n)= P ( n ) = u n = 3 n − 2 n u_{n} = 3^{n}- 2^{n} u n = 3 n − 2 n P ( n ) P(n) P ( n )
On a u 0 = 3 0 − 2 0 = 1 − 1 = 0 u_{0}= 3^0-2^0=1-1=0 u 0 = 3 0 − 2 0 = 1 − 1 = 0 u 1 = 3 1 − 2 1 = 3 − 2 = 1 u_{1} = 3^{1}-2^{1} = 3-2=1 u 1 = 3 1 − 2 1 = 3 − 2 = 1 P ( 0 ) P(0) P ( 0 ) P ( 1 ) P(1) P ( 1 )
Soit n ∈ N n\in\mathbb{N} n ∈ N P ( n ) P(n) P ( n ) P ( n + 1 ) P(n+1) P ( n + 1 ) P ( n + 2 ) P(n+2) P ( n + 2 )
On au n + 2 = 5 u n + 1 + 6 u n u_{n+2} = 5u_{n+1} + 6u_{n} u n + 2 = 5 u n + 1 + 6 u n = 5 ( 3 n + 1 − 2 n + 1 ) − 6 ( 3 n − 2 n ) \qquad \qquad \qquad \qquad \quad= 5(3^{n+1} - 2^{n+1}) - 6(3^{n} - 2^{n}) = 5 ( 3 n + 1 − 2 n + 1 ) − 6 ( 3 n − 2 n ) = 3 n ( 15 − 6 ) − 2 n ( 10 − 6 ) \qquad \qquad \qquad \quad= 3^{n}(15 - 6) - 2^{n}(10-6) = 3 n ( 15 − 6 ) − 2 n ( 10 − 6 ) = 3 n + 2 − 2 n + 2 \qquad= 3^{n+2} - 2^{n+2} = 3 n + 2 − 2 n + 2 P ( n + 2 ) P(n+2) P ( n + 2 )
Donc ∀ n ∈ N , u n = 3 n − 2 n \forall n \in \mathbb{N}, u_{n}=3^{n} - 2^{n} ∀ n ∈ N , u n = 3 n − 2 n
