Inductie matematica exercitiu rezolvat 1

$Cunoastem \ ca:$

$P_{n} = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ... + n(n + 1)(n + 2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$

$Demostrati \ ca \ daca:$

$P_{(n)} Adevarat \ =>$

$P_{(n)} -> P_{(n + 1)} \ Adevarat$

Rezolvare

$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ... + n(n + 1)(n + 2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$

$pentru \ n = 1 => P_{1} = 1 \cdot 2 \cdot 3 = \frac{1(1 + 1)(1 + 2)(1 + 3)}{4} => 6 = \frac{2 \cdot 3 \cdot 4}{4} => 6 = 6 \ Adevarat$

$presupunem \ ca:$

$P_{n} = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ... + n(n + 1)(n + 2) = \frac{n(n + 1)(n + 2)(n + 3)}{4} \ Adevarat$

$si \ trebuie \ sa \ demonstram \ ca:$

$P_{(n)} -> P_{(n + 1)}$

$P_{(n+1)} = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ... + n(n + 1)(n + 2) + (n + 1)(n + 2)(n + 3) = \frac{(n + 1)(n + 2)(n + 3)(n + 4)}{4} =>$

$si \ cum: \ 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ... + n(n + 1)(n + 2) = P_{(n)} = Adevarat =>$

$P_{(n+1)} = \frac{n(n + 1)(n + 2)(n + 3)}{4} + (n + 1)(n + 2)(n + 3) = \frac{(n + 1)(n + 2)(n + 3)(n + 4)}{4} =>$

$aducem \ la \ acelasi \ numitor \ 4 \ si \ eliminam \ numitorul \ si \ rezulta:$

$P_{(n+1)} = n(n + 1)(n + 2)(n + 3) + 4(n + 1)(n + 2)(n + 3) = (n + 1)(n + 2)(n + 3)(n + 4) =>$

$dam \ factor \ comun \ in \ stanga \ egalitatii \ pe: \ (n + 1)(n + 2)(n + 3) \ si \ rezulta:$

$P_{(n+1)} = (n + 1)(n + 2)(n + 3)(n + 4) = (n + 1)(n + 2)(n + 3)(n + 4) =>$

$P_{(n+1)} = egalitatea \ este \ Adevarata =>$

$P_{(n)} -> P_{(n + 1)} \ Adevarat$