ramadan mabrouk pour tt le monde
jé keke exo a vous proprsé
EX1
pr tt n £ IN on pose An= 3^(2n)-2^n Bn=2^(6n+3)+3^(2n+1)
An+1=2An+7.3^(2n) Bn+1=9Bn+55.2^(6n+3)
montre ke (An divise 7 ==>An+1 divise 7)é ( Bn divise 7 ==> Bn+1divise 7)
EX2
on a Sn=1^3+2^3+3^3+...+n^3 é Tn=1^2+2^2+3^2+...+n^2
1/ pr tt n £IN* montre ke Sn=[n(n+1)/2]^2==>Sn+1=[(n+1)(n+2)/2]^2
2/ dédui ke pr tt n £IN* Sn=[n(n+1)/2]^2
3/montre ke kelke soi n £IN* Tn=n(n+1)(2n+1)/6
Merci davance