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Pe: 2012-12-11T09:11:31+02:00 In: In:

inductie

am si eu o problema in legatura cu un exercitiu :

sa se arate ca pt orice numar n apartine N are loc relatia :

n la puterea a3a + 5n se divide cu 6

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2 raspunsuri

  1. Zeus
    veteran (III)

    n*(n^2+5) % 6=n*(n^2-1)%6=(n-1)*n*(n+1)%6=0 deoarece sunt 3 termeni consecutivi. Reamintesc ca 2 termeni consecutivi => impartire exacta a produsului lor la 2 iar 3 termeni consecutivi => impartire exacta a produsului lor la 6.

  2. ali
    maestru (V)

    Rezolvarea completa prin inductie ar fi:
    \begin{array}{l} {d_n}:{n^3} + 5n \vdots 6\\ \left. 1 \right){d_1} = 1 + 5 \vdots 6\\ {d_2} = 8 + 10 \vdots 6\\ \left. 2 \right){d_k}:{k^3} + 5k \to A\\ \left. 3 \right){d_k} \to {d_{k + 1}} \to A,\,{d_{k + 1}}:{\left( {k + 1} \right)^3} + 5\left( {k + 1} \right) \to A\\ {d_{k + 1}} = {\left( {k + 1} \right)^3} + 5\left( {k + 1} \right) = {k^3} + 3{k^2} + 3k + 1 + 5k + 5 = \underbrace {{k^3} + 5k}_{{d_k}} + 3{k^2} + 3k + 6 \vdots 6\\ \Leftrightarrow 3{k^2} + 3k + 6 \vdots 6 \Leftrightarrow 3{k^2} + 3k + 6 \vdots 2 \wedge 3\\ 3{k^2} + 3k + 6 = 3\left( {{k^2} + k + 2} \right) \vdots 3\\ {\rm{Demonstram ca }}{k^2} + k + 2 \vdots 2\\ \bullet k = 2p \Rightarrow 4{p^2} + 2p + 2 = 2\left( {2{p^2} + p + 1} \right) \vdots 2\\ \bullet k = 2p + 1 \Rightarrow {\left( {2p + 1} \right)^2} + 2p + 1 + 2 = 4{p^2} + 4p + 1 + 2p + 1 + 2 = 4{p^2} + 6p + 4\\ = 2\left( {2{p^2} + 3p + 2} \right) \vdots 2\\ \Rightarrow {d_{k + 1}} \vdots 3 \wedge 2 \Rightarrow {d_{k + 1}} \vdots 3 \times 2 = 6,\,\,\left( {2,3} \right) = 1. \end{array}

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