Ensino Superior ⇒ Sequência de Fibonacci Tópico resolvido

[theblackmamba]

Prove que:

[tex3]{n\choose n}F_0+{n\choose n-1}F_1+{n\choose n-2}F_2+...+{n\choose 0}F_n=F_{2n}[/tex3]

onde [tex3]F_n[/tex3] denota a sequência de Fibonacci.

[jedi]

do triangulo de pascal temos que

[tex3]{n\choose k}={n-1 \choose k-1}+{n-1 \choose k}[/tex3]

então podemos dizer que

[tex3]{n\choose n}F_0+{n\choose n-1}F_1+{n\choose n-2}F_2+...+{n\choose 0}F_n=[/tex3]

[tex3]{n-1\choose n-1}F_0+\left({n-1\choose n-1}+{n-1\choose n-2}\right)F_1+\left({n-1\choose n-2}+{n\choose n-3}\right)F_2+...+{n-1\choose 0}F_n=[/tex3]

[tex3]{n-1\choose n-1}(F_0+F_1)+{n-1\choose n-2}(F_1+F_2)+{n\choose n-3}(F_2+F_3)+...+{n-1\choose 0}(F_{n-1}+F_{n})=[/tex3]

[tex3]{n-1\choose n-1}F_2+{n-1\choose n-2}F_3+{n\choose n-3}F_4+...+{n-1\choose 0}F_{n+1}=[/tex3]

[tex3]{n-2\choose n-2}F_2+\left({n-2\choose n-2}+{n-2\choose n-3}\right)F_3+\left({n\choose n-3}+{n-2\choose n-4}\right)F_4+...+{n-1\choose 0}F_{n+1}=[/tex3]

[tex3]{n-2\choose n-2}(F_2+F_3)+{n-2\choose n-3}(F_3+F_4)+{n-2\choose n-4}(F_4+F_5)+...+{n-1\choose 0}(F_n+F_{n+1})=[/tex3]

[tex3]{n-2\choose n-2}F_4+{n-2\choose n-3}F_5+{n-2\choose n-4}F_6+...+{n-1\choose 0}F_{n+2}=[/tex3]

[tex3]\vdots[/tex3]

[tex3]{n-n\choose n-n}F_{n+n}=F_{2n}[/tex3]