الأولى باكالوريا العلوم التجريبية – خيار فرنسي
Soit (un)n∈N(u_n)_{n\in\N}(un)n∈N la suite numérique définie par son terme général : un=2n+1n+3u_n=\frac{2n+1}{n+3}un=n+32n+1
u0=2×0+10+3=13u_{0}=\dfrac{2\times0+1}{0+3}=\dfrac13u0=0+32×0+1=31
u1=2×1+11+3=34u_{1}=\dfrac{2\times1+1}{1+3}=\dfrac34u1=1+32×1+1=43
u20=2×20+120+3=4123u_{20}=\dfrac{2\times20+1}{20+3}=\dfrac{41}{23}u20=20+32×20+1=2341
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=n+1+32(n+1)+1=n+42n+3
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=n−1+32(n−1)+1=n+22n−1