[Citat] cum facem aici? as vrea dac se poate detaliat...pls multumesc. |
presupunem P(k) adevarat, adica 1+a+a^2+....a^k=[1-a^(k+1)]/(1-a)
calculam P(k+1) =>[ 1+a+a^2+......+a^k ] +a^(k+1)==[1-a^(k+1)]/(1-a) +a^(k+1)= aducem la acelasi numitor fractiile si obtinem = [ 1-a^(k+2)]/(1-a) => ca P(k) impilca P(k+1) adevarat=> P(n) adevarat