The Fibonacci Sequence is defined by \(F_0 = 1\),\(F_1 =1\) and \(F_n=F_{n-1}+F_{n-2}\). Prove that \(7{F_{n+2}^3}-{F_n^3}-{F_{n+1}^3}\) is divisible by \(F_{n+3}\).

This a part of my set NMTC 2nd Level (Junior) held in 2014.

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TopNewestLet us take \(F_{n}=x\) and \(F_{n+1}=y \Rightarrow\ F_{n+2}=x+y \Rightarrow\ F_{n+3}=x+2y.\)Now,let us expand the given expression:\(: 7F_{n+2}^{3}-F_{n}^{3}-F_{n+1}^{3}\) in terms of \(x\) and \(y\).We get\(: 7(x+y)^{3}-x^{3}-y^{3}\).Simplifying that,we get\(: 6x^{3}+6y^{3}+21xy(x+y).\)Taking \((x+y)\) common we get\(: (x+y)(6x^{2}-6xy+6y^{2}+21xy) =(x+y)(6x^{2}+15xy+6y^{2}) =(x+y)(x+2y)(6x+3y).\)But,\((2y+x)=F_{n+3}.\)hence proved:):).

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@Krishna Ar @Siddharth G

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Perfect! I didn't go for

x,yand faced problems in factorizing. PS: Small Typo at the end.Log in to reply

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There is very little here to motivate a solution by induction. Knowing divisibility by \( F_n \) doesn't tell you anything about divisibility by \( F_{n+1} \).

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Absolutely right. This is wht I tried, but the \(X,Y\) substitution, Mindblowing :D

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How did you do?

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Not well, 1b 5a, b 6b and half of the 8th were good. I left 1a half done. How was your paper?

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Same problem with wet 3a. BTW are you sure that 6a is 'Yes'?

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Sigh, atleast you had one full question to your credit. I got the first one fully , messed up second one (after getting half of it), third one wasn't salubrious, (I tried both, got to almost the answers), 4th I didnt do, 5a I didnt know, 5 b I got it, 6- I wrote Yes and No alternatingly :P, 7th I almost got it but lost it due to calculation error (You wont believe I put 40 cubed= 16000 :( )..8th I am not sure of it's accuracy....

Now , you clearly know I sucked more than you did. :(

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