📝 Exercices de la série▼
Exercice 1Exercice 2Exercice 3Exercice 4Exercice 5Exercice 6Exercice 7Exercice 8Exercice 9Exercice 10Exercice 11Exercice 12

Exercice 1

Soit (un)n∈N(u_n)_{n\in\N} la suite numérique définie par son terme général : un=2n+1n+3u_n=\frac{2n+1}{n+3}

  1. calculer u0u_{0} , u1u_{1} , u20u_{20}
  2. calculer un+1u_{n+1} , un−1u_{n-1}
  3. calculer un+1−unu_{n+1}-u_{n}

    • u0=2×0+10+3=13u_{0}=\dfrac{2\times0+1}{0+3}=\dfrac13

    • u1=2×1+11+3=34u_{1}=\dfrac{2\times1+1}{1+3}=\dfrac34

    • u20=2×20+120+3=4123u_{20}=\dfrac{2\times20+1}{20+3}=\dfrac{41}{23}

    • un+1=2(n+1)+1n+1+3=2n+3n+4u_{n+1}=\dfrac{2(n+1)+1}{n+1+3}=\dfrac{2n+3}{n+4}

    • un−1=2(n−1)+1n−1+3=2n−1n+2u_{n-1}=\dfrac{2(n-1)+1}{n-1+3}=\dfrac{2n-1}{n+2}

un+1−un=2n+3n+4−2n+1n+3=(n+3)(2n+3)−(n+4)(2n+1)(n+4)(n+3)=5(n+4)(n+3)\begin{align*} u_{n+1}-u_{n} &=\dfrac{2n+3}{n+4}-\dfrac{2n+1}{n+3} \\ &=\dfrac{(n+3)(2n+3)-(n+4)(2n+1)}{(n+4)(n+3)} \\ &=\dfrac{5}{(n+4)(n+3)} \end{align*}