Suites numériques

Expression explicite d'une suite et calculs de ses premiers termes - Exercice 2

10 min
20
COMPETENCES‾  :  Calculer.{\color{red}\underline{COMPETENCES}\;:\;Calculer.}
Question 1
Soit nn un entier naturel.
Calculer les trois premiers termes de chacune des suites suivantes :

un=5n2−7n−3u_{n} =5n^{2} -7n-3

Correction
u0=5×02−7×0−3u_{0} =5\times 0^{2} -7\times 0-3 donc
u0=−3u_{0} =-3

u1=5×12−7×1−3u_{1} =5\times 1^{2} -7\times 1-3 donc
u1=−5u_{1} =-5

u2=5×22−7×2−3u_{2} =5\times 2^{2} -7\times 2-3 donc
u2=3u_{2} =3
Question 2

un=−6+2n5n+3u_{n} =\frac{-6+2n}{5n+3}

Correction
u0=−6+02×0+3u_{0} =\frac{-6+0}{2\times 0+3} donc
u0=−63=−2u_{0} =\frac{-6}{3}=-2

u1=−6+2×15×1+3u_{1} =\frac{-6+2\times1}{5\times 1+3} donc
u1=−48=−12u_{1} =\frac{-4}{8}=-\frac{1}{2}

u2=−6+2×25×2+3u_{2} =\frac{-6+2\times2}{5\times 2+3} donc
u2=−213u_{2} =-\frac{2}{13}
Question 3

un=3n+1n+9u_{n} =\frac{3^{n+1} }{n+9}

Correction
u0=30+10+9u_{0} =\frac{3^{0+1} }{0+9} donc
u0=39=13u_{0} =\frac{3}{9}=\frac{1}{3}

u1=31+11+9u_{1} =\frac{3^{1+1} }{1+9} donc
u1=910u_{1} =\frac{9}{10}

u2=32+12+9u_{2} =\frac{3^{2+1} }{2+9} donc
u2=2711u_{2} =\frac{27}{11}
Question 4

un=4n+2nu_{n} =\sqrt{4^{n}+2n}

Correction
u0=40+2×0u_{0} =\sqrt{4^{0}+2\times 0} donc
u0=1=1u_{0} =\sqrt{1}=1
. On rappelle que 40=14^{0}=1
u1=41+2×1u_{1} =\sqrt{4^{1}+2\times 1} donc
u1=6u_{1} =\sqrt{6}

u2=42+2×2u_{2} =\sqrt{4^{2}+2\times 2} donc
u2=20u_{2} =\sqrt{20}