Voici un serpentin garni de cubes pour illustrer ce parcours
Pour l’utilisation de ce dessin, se reporter aux conditions définies dans
» identité du mois de janvier 2022 »
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A = ![\sum_{p=1}^{p=k}[p(p+2)]\](http://s0.wp.com/latex.php?latex=%5Csum_%7Bp%3D1%7D%5E%7Bp%3Dk%7D%5Bp%28p%2B2%29%5D%5C&bg=T&fg=000000&s=0 "\sum_{p=1}^{p=k}[p(p+2)]\")
![\sum_{p=1}^{p=k}[(p+1+n)(p+2+n)]\](http://s0.wp.com/latex.php?latex=%5Csum_%7Bp%3D1%7D%5E%7Bp%3Dk%7D%5B%28p%2B1%2Bn%29%28p%2B2%2Bn%29%5D%5C&bg=T&fg=000000&s=0 "\sum_{p=1}^{p=k}[(p+1+n)(p+2+n)]\")
( A )[ u + ( 12 + 6n )k + 2k² ] = ( B )( 7 + 2k )( k + 1 )
n = 0 – > u = 22
n = 1 – > u = 52
n = 2 – > u = 94
n = 3 – > u = 148
n = 4 – > u = 214
etc …
u prend les valeurs de la suite 
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A =
![\sum_{p=1}^{p=k}[(p+1+n)(p+3+n)]\](http://s0.wp.com/latex.php?latex=%5Csum_%7Bp%3D1%7D%5E%7Bp%3Dk%7D%5B%28p%2B1%2Bn%29%28p%2B3%2Bn%29%5D%5C&bg=T&fg=000000&s=0 "\sum_{p=1}^{p=k}[(p+1+n)(p+3+n)]\")
( A )[ v + ( 15 + 6n )k + 2k² ] = ( B )( 7 + 2k )( k + 1 )
n = 0 -> v = 31
n = 1 – > v = 67
n = 2 – > v = 115
n = 3 – > v = 175
n = 4 – > v = 247
etc …
v prend les valeurs de la suite 
