Parcial 2011 induccion. (2014)

Examen Español
Universidad Universidad Politécnica de Cataluña (UPC)
Grado Ingeniería Civil - 1º curso
Asignatura Fundamentos Matematicos
Año del apunte 2014
Páginas 7
Fecha de subida 03/06/2014
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Examen parcial 2011 induccion resuelto.

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❊♥❣✐♥②❡r✐❛ ❈✐✈✐❧✳ ❋♦♥❛♠❡♥ts ♠❛t❡♠àt✐❝s✳ ✷✷✴✵✾✴✷✵✶✶✳ • ❉❡♠♦str❡✉ ♣❡r ✐♥❞✉❝❝✐ó ❧❛ ♣r♦♣♦s✐❝✐ó✿ ∀n ∈ N, n ≥ 2, 122n−1 + 11n+1 és ✉♥ ♠ú❧t✐♣❧❡ ♥❛t✉r❛❧ ❞❡ ✶✸✸ [ n = 2 ] 122n−1 + 11n+1 = 123 + 113 = 1728 + 1331 = 3059 = 23 · 133 [n=m] ❙✉♣♦s❡♠ ❝❡rt q✉❡✿ 122m−1 + 11m+1 = 133k k ∈ N✱ m ≥ 2 ♣❡r ❛❧❣✉♥ ✭❍✐♣òt❡s✐ ❞✬■♥❞✉❝❝✐ó✱ ❍✳■✳✮ [ n=m+1 ] ❊♥s ♣r❡❣✉♥t❡♠✿ ➱s 122(m+1)−1 + 11(m+1)+1 ✉♥ ♠ú❧t✐♣❧❡ ♥❛t✉r❛❧ ❞❡ 133 ❄ ✱ és ❛ ❞✐r✱ és 122m+1 + 11m+2 ✉♥ ♠ú❧t✐♣❧❡ ♥❛t✉r❛❧ ❞❡ 133 ❄ ◆♦t❡♠ q✉❡ ❧❛ ❤✐♣òt❡s✐ ❞✬✐♥❞✉❝❝✐ó ❧❛ ♣♦❞❡♠ ❡s❝r✐✉r❡ ❝♦♠✿ 122m−1 = 133k − 11m+1 ♣❡r ❛❧❣✉♥ k∈N ✭✶✮ ▲❧❛✈♦rs✿ (1) 122m+1 + 11m+2 = 122 · 122m−1 + 11m+2 = (1) = 144(133k − 11m+1 ) + 11 · 11m+1 = 133 · (144k) + 11m+1 (−144 + 11) = = 133 · (144k) − 133 · 11m+1 = 133 · (144k − 11m+1 ) ❆r❛ ♥♦♠és q✉❡❞❛ ✈❡✉r❡ q✉❡ 144k − 11m+1 és ✉♥ ♥♦♠❜r❡ ♥❛t✉r❛❧✳ P❡rò ❛✐①ò és ❢à❝✐❧ ❛ ♣❛rt✐r ❞❡ ❧❛ ❍✳■✳✿ H.I.
144k − 11m+1 ≥ 133k − 11m+1 = 122m−1 + 11m+1 − 11m+1 = 122m−1 ∈ N , ✜♥❛❧✐t③❛♥t ♣❡r t❛♥t ❧❛ ❞❡♠♦str❛❝✐ó✳ ❊♥❣✐♥②❡r✐❛ ❈✐✈✐❧✳ ❋♦♥❛♠❡♥ts ♠❛t❡♠àt✐❝s✳ ✷✷✴✵✾✴✷✵✶✶✳ • ❉❡♠♦str❡✉ ♣❡r ✐♥❞✉❝❝✐ó ❧❛ ❞❡s✐❣✉❛❧t❛t✿ n √ 1 √ ≥ n ∀n ∈ N, n ≥ 2 k k=1 n 2 [n=2] k=1 √ ❥❛ q✉❡ √ 1 1 1 √ = √ =1+ √ ≥ 2 , 2 k k=1 k √ √ √ 1 2 · (1 + √ ) = 2 + 1 ≥ 2 = 2 · 2 2 m [n=m] ❙✉♣♦s❡♠ ❝❡rt q✉❡✿ k=1 [ n=m+1 ] √ ❡♥ s❡r √ 1 √ ≥ m k √ 2 ≥ 1 ( 2 ≈ 1.41) ✭❍✐♣òt❡s✐ ❞✬■♥❞✉❝❝✐ó✱ ❍✳■✳✮ ❊♥s ♣r❡❣✉♥t❡♠✿ m+1 k=1 √ 1 √ ≥ m+1 k ❄ P♦❞❡♠ ❡s❝r✐✉r❡✿ m+1 k=1 1 √ k m = k=1 H.I. √ 1 1 1 √ +√ ≥ = m+ √ m+1 m+1 k ✐ ✈♦❧❡♠ ✈❡✉r❡ q✉❡ ❛q✉❡st❛ ❞❛rr❡r❛ ❢r❛❝❝✐ó és ≥ √ m + 1✳ m(m + 1) + 1 √ , m+1 ✭✷✮ P❡rò ♥♦t❡♠ q✉❡✿ m(m + 1) + 1 √ √ ≥ m + 1 ⇐⇒ m(m + 1) + 1 ≥ m + 1 ⇐⇒ m(m + 1) ≥ m ⇐⇒ m+1 ⇐⇒ m(m + 1) ≥ m2 ⇐⇒ m2 + m = m2 ⇐⇒ m ≥ 0 ✭✸✮ ✭♦♥ s✬❤❛ t✐♥❣✉t ❡♥ ❝♦♠♣t❡ q✉❡ ❡❧s ♥♦♠❜r❡s ❞❡❧s q✉❛❧s ❛♣❛r❡✐① ❧✬❛rr❡❧ q✉❛❞r❛❞❛ só♥ ♥♦ ♥❡❣❛t✐✉s✮✳ P❡r t❛♥t✱ ❝♦♠ q✉❡ m ≥ 0✱ ❞❡ ✭✷✮ ✐ ✭✸✮ ❞❡❞✉ï♠ q✉❡ m+1 k=1 ✜♥❛❧✐t③❛♥t ❧❛ ❞❡♠♦str❛❝✐ó✳ √ 1 √ ≥ m+1 , k ❊♥❣✐♥②❡r✐❛ ❈✐✈✐❧✳ ❋♦♥❛♠❡♥ts ♠❛t❡♠àt✐❝s✳ ✷✷✴✵✾✴✷✵✶✶✳ • ❉❡♠♦str❡✉ ♣❡r ✐♥❞✉❝❝✐ó ❧❛ ✐❣✉❛❧t❛t✿ 2n (−1)k−1 = k k=1 2n [n=1] k=1 2 (−1)k−1 = k k=1 ❙✉♣♦s❡♠ ❝❡rt q✉❡✿ k=1 [ n=m+1 ] k=n+1 1 , k ∀n ∈ N∗ (−1)k−1 −1 1 =1+ = = k 2 2 2m [n=m] 2n (−1)k−1 = k 2m k=m+1 1 k 2 k=2 1 = k m≥1 ✱ 2n k=n+1 1 k ✭❍✐♣òt❡s✐ ❞✬■♥❞✉❝❝✐ó✱ ❍✳■✳✮ ❊♥s ♣r❡❣✉♥t❡♠✿ 2(m+1) k=1 (−1)k−1 = k 2(m+1) k=(m+1)+1 1 k ❄ ❉❡s❡♥✈♦❧✉♣❛r❡♠ ❡❧ ♠❡♠❜r❡ ❡sq✉❡rr❡ ✜♥s ❛ ♣♦❞❡r ❛♣❧✐❝❛r ❧❛ ❍✳■✳✱ ✐ ❛♠❜ ❧❛ ✐♥t❡♥❝✐ó ❞❡ ✈❡✉r❡ q✉❡ és ✐❣✉❛❧ ❛❧ ♠❡♠❜r❡ ❞r❡t✿ 2(m+1) k=1 (−1)k−1 k 2m+2 = k=1 2m = (−1)k−1 = k (−1)k−1 k=1 = 1 + m+1 2m+1 = k=m+2 ✭♦♥ s✬❤❛ ✉t✐❧✐t③❛t q✉❡ k 2m k=1 (−1)k−1 (−1)2m (−1)2m+1 + + = k 2m + 1 2m + 2 1 1 H.I.
+ − = 2m + 1 2m + 2 2m+1 k=m+2 1 1 − = k 2m + 2 1 1 + = k 2m + 2 2m+2 k=m+2 2m+1 2m k=m+1 1 + k k=m+2 2(m+1) 1 = k k=(m+1)+1 1 1 1 + − = k 2m + 1 2m + 2 1 1 − m + 1 2m + 2 = 1 k (−1)2m = 1, (−1)2m+1 = −1✱ ❥❛ q✉❡ 2m és ♣❛r❡❧❧ ✐ 2m + 1 és s❡♥❛r✮✳ ❊♥❣✐♥②❡r✐❛ ❈✐✈✐❧✳ ❋♦♥❛♠❡♥ts ♠❛t❡♠àt✐❝s✳ ✷✷✴✵✾✴✷✵✶✶✳ • ❉❡s❝r✐✉r❡ ❡❧ ❧❧♦❝ ❣❡♦♠ètr✐❝ ❞❡❧s ♣✉♥ts ❞❡❧ ❝♦♥❥✉♥t ❆ A = z ∈ C : |z + 2|2 ≤ 2Im(z) ❈♦♥s✐❞❡r❡♠ z = x + iy |z + 2|2 = |x + iy + 2|2 = ( (x + 2)2 + y 2 )2 = (x + 2)2 + y 2 ▲❧❛✈♦rs (x + 2)2 + y 2 ≤ 2y (x + 2)2 + y 2 − 2y ≤ 0 (x + 2)2 + (y − 1)2 ≤ 1 ❘❡s✉❧t❛t✿ ❆ és ❡❧ ❞✐s❝ ❞❡ ❝❡♥tr❡ ✭✲✷✱✶✮ ✐ r❛❞✐ ✶✳ ❊♥❣✐♥②❡r✐❛ ❈✐✈✐❧✳ ❋♦♥❛♠❡♥ts ♠❛t❡♠àt✐❝s✳ ✷✷✴✵✾✴✷✵✶✶✳ • ❉❡s❝r✐✉r❡ ❡❧ ❧❧♦❝ ❣❡♦♠ètr✐❝ ❞❡❧s ♣✉♥ts ❞❡❧ ❝♦♥❥✉♥t ❆ A = {z ∈ C : |z − 1| < |z − i|} ❈♦♥s✐❞❡r❡♠ z = x + iy |z − 1| = |x + iy − 1| = (x − 1)2 + y 2 |z − i| = |x + iy − i| = x2 + (y − 1)2 ▲❧❛✈♦rs |z − 1| < |z − i| (x − 1)2 + y 2 < x2 + (y − 1)2 x2 − 2x + 1 + y 2 < x2 + y 2 − 2y + 1 −2x < −2y x>y ❘❡s✉❧t❛t✿ ❆ és ❡❧ s❡♠✐♣❧à s♦t❛ ❧❛ r❡❝t❛ y=x s❡♥s❡ ✐♥❝❧✉♦r❡✲❧❛✳ ❊♥❣✐♥②❡r✐❛ ❈✐✈✐❧✳ ❋♦♥❛♠❡♥ts ♠❛t❡♠àt✐❝s✳ ✷✷✴✵✾✴✷✵✶✶✳ • ❉❡♠♦str❡✉ q✉❡ ♣❡r ❛ ❝❛❞❛ ♥♦♠❜r❡ r❡❛❧ x ❡①✐st❡✐① ✉♥ ❡♥t❡r ♣♦s✐t✐✉ n t❛❧ q✉❡ n > x✳ ❉❡♠♦str❛❝✐ó ♣❡r r❡❞✉❝❝✐ó ❛ ❧✬❛❜s✉r❞✳ ❙✐ ♥♦ ❢♦s ❛✐①í✱ ❡①✐st✐r✐❛ ✉♥ x q✉❡ s❡r✐❛ ❝♦t❛ s✉♣❡r✐♦r ❞❡ Z+ ✱ ❝♦♥tr❛❞✐❝❝✐ó ❛♠❜ ❡❧ ❢❡t q✉❡ ❡❧ + ❝♦♥❥✉♥t Z ❞❡❧s ♥♦♠❜r❡s ❡♥t❡rs ♣♦s✐t✐✉s ♥♦ ❡stà ✜t❛t s✉♣❡r✐♦r♠❡♥t✳ ❊♥❣✐♥②❡r✐❛ ❈✐✈✐❧✳ ❋♦♥❛♠❡♥ts ♠❛t❡♠àt✐❝s✳ ✷✷✴✵✾✴✷✵✶✶✳ • ❉❡♠♦str❡✉ q✉❡ s✐ A P❡r ❛ ❞❡♠♦str❛r q✉❡ és ✉♥ ❝♦♥❥✉♥t q✉❛❧s❡✈♦❧✱ ∅⊆A ∅ ⊆ A✳ x ∈ ∅ =⇒ x ∈ A✱ ♣❡r ❛ q✉❛❧s❡✈♦❧ ♦❜❥❡❝t❡ ✏ x✑✳ q✉❛❧s❡✈♦❧ x✮✱ ❞❡ ♠❛♥❡r❛ q✉❡✱ ♣❡r ❞❡✜♥✐❝✐ó✱ ❧❛ ✐♠♣❧✐❝❛❝✐ó ❝❛❧ ✈❡✉r❡ q✉❡ P❡rò x ∈ ∅ és s❡♠♣r❡ ❢❛❧s ✭♣❡r ❛ x ∈ ∅ =⇒ x ∈ A és ✈❡rt❛❞❡r❛✳ ❯♥ r❛♦♥❛♠❡♥t ❛❧t❡r♥❛t✐✉ és✿ s✐ ∅ A ❛✐①ò ♥♦ és ♣♦ss✐❜❧❡ ♣❡rq✉è ❡❧ ❝♦♥❥✉♥t ❜✉✐t ∅ ♥♦ té ❡❧❡♠❡♥ts✳ ∅ A✱ ♣❡rò P❡r t❛♥t✱ ❡s ❞❡❞✉❡✐① q✉❡ ∅ ⊆ A✳ ❤✐ ❤❛✉r✐❛ ❛❧❣✉♥ ❡❧❡♠❡♥t ❞❡ q✉❡ ♥♦ s❡r✐❛ ❞❡ ...