[tex]\displaystyle \sf 4+10+16+...+(6n-2) = n(3n+1) \ ;\ \forall \ n\ge 1\\\\ n = 1 \to 4 = .(3.1+1) = 4 \checkmark \\\\ n=2 \to 4+10 = 2(3.2+1) =14 \checkmark \\\\ n = 3 \to 4+10+16 = 3(3.3+1) = 30\checkmark \\\\\\ \underline{\text{Hip\'otese}} :\ \text{V\'alido }\forall\ k\ge 1 \\\\\ 4+10+16+...+(6k-2) = k(3k+1) \ \ ,\ \forall \ k \geq 1 \\\\\\\ \underline{\tex{Tese}} : \text{V\'alido }\forall \ (k+1) \ge 1[/tex]
[tex]\displaystyle \sf \underbrace{\sf 4+10+16+...+(6k-2)}_{k(3k+1)}+[6(k+1)-2] \overset{?}{=} (k+1)[3(k+1)+1] \\\\\\ k(3k+1) +6k+6-2 = (k+1)(3k+3+1) \\\\ k(3k+1)+6k+4 \overset{?}{=} (k+1)(3k+4)\\\\ k(3k+1+3)+6k+4-3k \overset{?}{=} (k+1)(3k+4) \\\\ k(3k+4)+6k+4-3k \overset{?}{=} (k+1)(3k+4) \\\\[/tex]
[tex]\displaystyle \sf k(3k+4)+3k+4 \overset{?}{=} (k+1)(3k+4) \\\\ \Large\underbrace{\boxed{\sf (k+1)(3k+4) = (k+1)(3k+4)}}_{\displaystyle \text{C.Q.D}} \ \checkmark \\\\\\[/tex]
[tex]\displaystyle \sf \boxed{\begin{matrix}\text{Ent\~ao de fato \'e v\'alido } : \\\\ \sf 4+10+16+...+(6n-2)=n(3n+1) \ , \ \forall \ n\geq 1 \end{matrix}\ }\checkmark[/tex]
Demostraremos usando o princípio da indução finita. Para isso seguiremos alguns passos:
[tex]\sf (6.1-2)=1(3.1+1) \Rightarrow4=4~~\checkmark[/tex]
[tex]\sf 4 + 10 + 16 + ... + (6k - 2) = k(3k + 1)[/tex]
[tex]\sf 4 + 10 + 16 + ... + (6k - 2)+(6(k+1)-2) = (k+1)(3(k+1) + 1)[/tex]
Dá hipótese, obtemos
[tex]\sf k(3k+1)+(6(k+1)-2) = (k+1)(3(k+1) + 1)[/tex]
Manipulando o membro esquerdo, obtemos
[tex]\sf k(3k+1)+(6(k+1)-2) \\ \\ \sf = k(3k+1)+(6k+6-2)\\ \\ \sf =3k^2+k+6k+4\\ \\ \sf=3(k^2+2k)+k+4\\ \\ \sf =3(k+1)^2+k+4-3\\ \\ \sf=3(k+1)^2+(k+1)\\ \\ \sf=(k+1)(3(k+1)+1)[/tex]
Com isso, provamos que é válido para o sucessor de k.
c.q.d
Copyright © 2024 ELIBRARY.TIPS - All rights reserved.
Lista de comentários
[tex]\displaystyle \sf 4+10+16+...+(6n-2) = n(3n+1) \ ;\ \forall \ n\ge 1\\\\ n = 1 \to 4 = .(3.1+1) = 4 \checkmark \\\\ n=2 \to 4+10 = 2(3.2+1) =14 \checkmark \\\\ n = 3 \to 4+10+16 = 3(3.3+1) = 30\checkmark \\\\\\ \underline{\text{Hip\'otese}} :\ \text{V\'alido }\forall\ k\ge 1 \\\\\ 4+10+16+...+(6k-2) = k(3k+1) \ \ ,\ \forall \ k \geq 1 \\\\\\\ \underline{\tex{Tese}} : \text{V\'alido }\forall \ (k+1) \ge 1[/tex]
[tex]\displaystyle \sf \underbrace{\sf 4+10+16+...+(6k-2)}_{k(3k+1)}+[6(k+1)-2] \overset{?}{=} (k+1)[3(k+1)+1] \\\\\\ k(3k+1) +6k+6-2 = (k+1)(3k+3+1) \\\\ k(3k+1)+6k+4 \overset{?}{=} (k+1)(3k+4)\\\\ k(3k+1+3)+6k+4-3k \overset{?}{=} (k+1)(3k+4) \\\\ k(3k+4)+6k+4-3k \overset{?}{=} (k+1)(3k+4) \\\\[/tex]
[tex]\displaystyle \sf k(3k+4)+3k+4 \overset{?}{=} (k+1)(3k+4) \\\\ \Large\underbrace{\boxed{\sf (k+1)(3k+4) = (k+1)(3k+4)}}_{\displaystyle \text{C.Q.D}} \ \checkmark \\\\\\[/tex]
[tex]\displaystyle \sf \boxed{\begin{matrix}\text{Ent\~ao de fato \'e v\'alido } : \\\\ \sf 4+10+16+...+(6n-2)=n(3n+1) \ , \ \forall \ n\geq 1 \end{matrix}\ }\checkmark[/tex]
Resolução
Demostraremos usando o princípio da indução finita. Para isso seguiremos alguns passos:
[tex]\sf (6.1-2)=1(3.1+1) \Rightarrow4=4~~\checkmark[/tex]
[tex]\sf 4 + 10 + 16 + ... + (6k - 2) = k(3k + 1)[/tex]
[tex]\sf 4 + 10 + 16 + ... + (6k - 2)+(6(k+1)-2) = (k+1)(3(k+1) + 1)[/tex]
Dá hipótese, obtemos
[tex]\sf k(3k+1)+(6(k+1)-2) = (k+1)(3(k+1) + 1)[/tex]
Manipulando o membro esquerdo, obtemos
[tex]\sf k(3k+1)+(6(k+1)-2) \\ \\ \sf = k(3k+1)+(6k+6-2)\\ \\ \sf =3k^2+k+6k+4\\ \\ \sf=3(k^2+2k)+k+4\\ \\ \sf =3(k+1)^2+k+4-3\\ \\ \sf=3(k+1)^2+(k+1)\\ \\ \sf=(k+1)(3(k+1)+1)[/tex]
Com isso, provamos que é válido para o sucessor de k.
c.q.d