G greg1313 Dec 2016 295 161 Earth Oct 27, 2019 #1 For \(\displaystyle \ n \ge 3, \ \ \) prove that \(\displaystyle F_n + F_{n + 2} \ge 2F_{n + 1} + 1\) Last edited by a moderator: Oct 29, 2019

For \(\displaystyle \ n \ge 3, \ \ \) prove that \(\displaystyle F_n + F_{n + 2} \ge 2F_{n + 1} + 1\)

P Plato MHF Helper Aug 2006 22,469 8,640 Oct 27, 2019 #2 greg1313 said: For \(\displaystyle \ n \ge 3, \ \ \) prove that \(\displaystyle F_n \ + \ F_{n + 2} \ \ge \ 2F_{n + 1} \ + \ 1\) Click to expand... Hint: Spoiler I hope that you know that $F_1=1,~F_2=1~\&~n\ge 3 \Rightarrow F_{n}=F_{n-1}+F_{n-2}$ Use induction: is this true for the base case, $n=3~?$ Suppose that it is true for $n=K>3$. then look at $F_{K+1}=F_{(K+1)-1}+F_{(K+1)-2}$ Last edited by a moderator: Oct 29, 2019

greg1313 said: For \(\displaystyle \ n \ge 3, \ \ \) prove that \(\displaystyle F_n \ + \ F_{n + 2} \ \ge \ 2F_{n + 1} \ + \ 1\) Click to expand... Hint: Spoiler I hope that you know that $F_1=1,~F_2=1~\&~n\ge 3 \Rightarrow F_{n}=F_{n-1}+F_{n-2}$ Use induction: is this true for the base case, $n=3~?$ Suppose that it is true for $n=K>3$. then look at $F_{K+1}=F_{(K+1)-1}+F_{(K+1)-2}$

topsquark Forum Staff Jan 2006 11,565 3,453 Wellsville, NY Oct 29, 2019 #3 Plato: This is the Challenge Forum. He knows how to solve it. -Dan