Question

Exercitiul 97 punctul d

Answer 1

cf pct c)
X(a) *X(b)=X(a+b+ab)
dar a+b+ab=a+b+ab+1-1=(a+1)(b+1)-1
Obs. se poate arata ca acesta lege este asociativa ( *)
vezui legi de compozitie ; nu ungim demomostratia

atunci
X(a).*X(b)= X[(a+1)(b+1)-1]
si
[X(a)*X(b)]*X(c)= X[(a+1)(b+1)-1]*X9c) =X {[(a+1)(b+1)-1]+1}(b+1)-1}=
x[(a+1)(b+1)(c+1)=1]
pt a=1, b=2,c=3
X (1)*X(2)*X(3)= X (2*3*4-1)=X ( 1*2*3*4-1)=X(4!-1)
 presupunem ca
X(1)* X(2)...* X (n)= X [(n+1)!-1]     (1)

 verificam pt n=1  (dar deja a fost verifica pt n=3)

 X(1)* X(2)= X [(1+1) (2+1)-1]=X(2*3-1)=X(3!-1) adevarat


Presupunem adevarata pt. n, cu forma (1)

calculam pt n+1
X(1)* X(2)...* X (n)* X (n+1) = X[(n+1)!-1]* X (n+1)

am folosit Pn adevarat pt primii n factori
 Relatia devine :
 X {[(n+1)!-1+1]*(n+1+1)-1}= X{(n+1)!* (n+2)-1]= X [(n+2)!-1]
deci
Pn⇒Pn+1
deci relatia presupusa este adevarata ∀n∈N
 atunci
X(1) *X9(2)*...* X(2007)= X [ 2007+1)!-1]= X ( 2008!-1), cerinta