Subversion Repositories wimsdev

% traduzione locale di H5/analysis/suites.fr -- suites_rel_rec_par
\title{Ricorrenza}
\language{it}
\range{-5..5}
\author{Frédéric, Pitoun}
\email{frederic.pitoun@free.fr}
\computeanswer{yes}
\format{html}
\precision{1000}

\integer{a=randint(-1,1)*randint(1..5)}
\integer{b=randint(1..5)}
\integer{c=randint(-1,1)*randint(1..5)}

\integer{u0=randint(-5..5)}
\function{f=maxima(\b*n+\c)}
\integer{p=\confparm1= ? 2:\confparm1}
\integer{n=randint(2..\p)}

\integer{rep1=\a*\u0+\b*0+\c}
\integer{rep2=\a*\rep1+\b*1+\c}

\function{fonc=maxima(\a*x+\b*y+\c)}
\text{fonc=wims(replace internal y by n in \fonc)}
\text{fonc=wims(replace internal x by u_n in \fonc)}

\text{rel=texmath(\fonc)}







\statement{
Considerare la sequenza \((u_n)) definita ricorsivamente ponendo \(u_0=\u0) e 
<p class="wimscenter">
\(u_{n+1}=\rel)
</p>
Calcolare \(u_1) e \(u_2) .<br/>
<p class="wimscenter">
Il termine \(u_1\)  vale \(\;\;\) \embed{reply 1,5}
</p>
<p class="wimscenter">
Il termine \(u_2\)  vale \(\;\;\) \embed{reply 2,5}
</p>
}


\answer{Risposta u_1}{\rep1}{type=default}
\answer{Risposta u_2}{\rep2}{type=default}