Suites numériques

Expression explicite d'une suite et calculs de ses premiers termes - Exercice 3

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=13n26n+1u_{n} =\frac{1}{3}n^{2} -6n+1

Correction
u0=13×026×0+1u_{0} =\frac{1}{3}\times 0^{2} -6\times 0+1 donc
u0=1u_{0} =1

u1=13×126×1+1u_{1} =\frac{1}{3}\times 1^{2} -6\times 1+1
u1=136+1u_{1} =\frac{1}{3} -6+1
u1=136×31×3+1×31×3u_{1} =\frac{1}{3} -\frac{6\times 3}{1\times 3}+\frac{1\times 3}{1\times 3}
u1=118+33u_{1} =\frac{1-18+3}{3} donc
u1=143u_{1} =-\frac{14}{3}

u2=13×226×2+1u_{2} =\frac{1}{3}\times 2^{2} -6\times 2+1
u2=4312+1u_{2} =\frac{4}{3} -12+1
u2=4312×31×3+1×31×3u_{2} =\frac{4}{3} -\frac{12\times 3}{1\times 3}+\frac{1\times 3}{1\times 3}
u2=436+33u_{2} =\frac{4-36+3}{3} donc
u2=293u_{2} =-\frac{29}{3}
Question 2

un=n+64n8u_{n} =\frac{n+6}{4n-8}

Correction
u0=0+64×08u_{0} =\frac{0+6}{4\times 0-8} donc
u0=68=34u_{0} =-\frac{6}{8}=-\frac{3}{4}

u1=1+64×18u_{1} =\frac{1+6}{4\times 1-8} donc
u1=74u_{1} =-\frac{7}{4}

u2=2+64×28u_{2} =\frac{2+6}{4\times 2-8} donc
u2=80=0u_{2} =-\frac{8}{0}=0
Question 3

un=5n24nu_{n} =\frac{5n-2}{4^{n}}

Correction
u0=5×0240u_{0} =\frac{5\times 0-2}{4^{0} } donc
u0=2u_{0} =-2
. On rappelle que 40=14^{0}=1
u1=5×1241u_{1} =\frac{5\times 1-2}{4^{1} } donc
u1=34u_{1} =\frac{3}{4}

u2=5×2242u_{2} =\frac{5\times 2-2}{4^{2} } donc
u2=816=12u_{2} =\frac{8}{16}=\frac{1}{2}
Question 4

un=14n+7u_{n} =\frac{1}{4}\sqrt{n+7}

Correction
u0=142×0+7u_{0} =\frac{1}{4}\sqrt{2\times 0+7} donc
u0=74u_{0} =\frac{\sqrt{7}}{4}

u1=142×1+7u_{1} =\frac{1}{4}\sqrt{2\times 1+7} donc
u1=94=34u_{1} =\frac{\sqrt{9}}{4} =\frac{3}{4}

u2=142×2+7u_{2} =\frac{1}{4}\sqrt{2\times 2+7} donc
u2=114u_{2} =\frac{\sqrt{11}}{4}