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sj_elf_ro
Pe: 2012-12-12T17:00:46+02:00 In: In:

divizibilitate

Sa se arate ca orice numar natural n>=2 exista numerele naturale distincte a1, a2, …, an divizibile cu 3 astfel incat 1\2=1\a1+a\a2+…+1\an

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  1. mathmagix

    P(2): \frac{1}{2}= \frac{1}{3}+\frac{1}{6} \\ \text{deci} a_1=3,a_2=6 \\ \text{Pp } P(k) \text{ adevarata}: \frac{1}{2}= \frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_k} \\ \text{Demonstram } P(k+1) \text{ adevarata}: \frac{1}{2}= \frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_{k+1}} \\ \frac{1}{2}=\frac{1}{3}+\frac{1}{6}=\frac{1}{3}+\frac{1}{3}\cdot \frac{1}{2}=\frac{1}{3}+\frac{1}{3}\cdot(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_k}) \\ \text{Astfel am pus in evidenta} k+1 \text{numere}:a'_1=3,a'_2=3a_1,...,a'_{k+1}=3a_k
    Se observa ca acestea sunt distincte si divizibilw cu 3.

  2. sj_elf_ro

    Va multumesc!!!

  4. PhantomR
    expert (VI)

    mathmagixsj_elf_ro pentru postarea problemei.

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