tarajo3

[email]alloirat_121@hotmail.fr[/email]

pour n=1 evident pn vrais
soit n£N*
on supose pn vrais et on montre que p(n+1) vrais

1/1*2*3 + 1/2*3*4+-----+ 1/ n(n+1)(n+2)+ 1/(n+1)(n+2)(n+3)=(n+1)(n+4)/4(n+2)(n+3)

p(n+1)=p(n)+1/(n+1)(n+2)(n+3)

=> p(n+1)=n(n+3)/4(n+1)(n+2)+ 1/(n+1)(n+2)(n+3)
=> n(n+3)²+4/4(n+1)(n+2)(n+3)
=>n^3+9n+6n²+4/4(n+1)(n+2)(n+3)
=>(n+1)(n²+5n+4)/4(n+1)(n+2)(n+3)
n²+5n+4 determinant delta=3>0 calculex1&x2 =>n²+5n+4= (n+4)(+1)
p(n+1)= (n+1)(n+4)(n+1)/4(n+1)(n+2)(n+3)
donc pn+1= (n+4)(n+1)/4(n+2)(n+3)
hc