Démontrer par récurrence que :

Bonsoir,

1) transformer en fractions simples:
1/(k(k+1)(k+2)=  a/k+b/(k+1)+c/(k+2) = 1/(2k) -1/(k+1)+1/(2(k+2))
On pose H(n)= 1+1/2+1/3+1/4+1/5+...+1/n
H(n+1)=H(n)+1/(n+1)

s=somme(de k=1 à n) (1/(k(k+1)(k+2))
s=1/2 H(n)  - (H(n+1) -1) +1/2( H(n+2)-(1+1/2))
=1/2H(n)-(H(n)+1/(n+1)-1)+1/2(H(n)+1/(n+1)+1/(n+2)-3/2)
=-1/(n+1)+1+1/(2(n+1))+1/(2(n+2))-3/4
=-1/(2(n+1)) +1/(2(n+2))+1/4
=(n²+3n)/(4(n+1)(n+2))=n(n+3)/(4(n+1)(n+2))