a) initialisation :
vrai pour n=1
b)heredite :
on suppose
A =(1^3)+(2^3)+...+n^3 = [n²(n+1)²]/4
on calcule B =(1^3)+(2^3)+...+n^3 +(n+1)^3 (et on aimerait montrer que ca vaut
C =[(n+1)²(n+2)²]/4
B = A + (n+1)^3
= [n²(n+1)²]/4  + n^3 + 3n² +3n +1
= n²(n²+2n+1)/4 +(4n^3 + 12n² +12n +4 )/4
= ( n^4 +2n^3+n² + 4n^3 + 12n² +12n +4 ) / 4
= (n^4 +6n^3+ 13n² +12n +4 ) / 4
= C   ( il suffit de developper C pour le verifier )
c)conclusion ...
bonne comprehension
spmtb