Nu pot sa-mi dau seama cum se calculeaza o astfel de cumpunere:
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Daca f:R->R; f(x)=a*x+b cu a; b numere reale , a nenul atunci se demonstreaza prin inductie ca f compusa de n ori cu ea insasi este egala cu (a^n*)x+b(1+a+a^2+…+a^(n-1)) pentru orice n>1 natural