The Fibonacci Sequence is defined by F0=1F_0 = 1F0=1,F1=1F_1 =1F1=1 and Fn=Fn−1+Fn−2F_n=F_{n-1}+F_{n-2}Fn=Fn−1+Fn−2. Prove that 7Fn+23−Fn3−Fn+137{F_{n+2}^3}-{F_n^3}-{F_{n+1}^3}7Fn+23−Fn3−Fn+13 is divisible by Fn+3F_{n+3}Fn+3.
This a part of my set NMTC 2nd Level (Junior) held in 2014.
Note by Siddharth G 6 years, 7 months ago
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Let us take Fn=xF_{n}=xFn=x and Fn+1=y⇒ Fn+2=x+y⇒ Fn+3=x+2y.F_{n+1}=y \Rightarrow\ F_{n+2}=x+y \Rightarrow\ F_{n+3}=x+2y.Fn+1=y⇒ Fn+2=x+y⇒ Fn+3=x+2y.Now,let us expand the given expression::7Fn+23−Fn3−Fn+13: 7F_{n+2}^{3}-F_{n}^{3}-F_{n+1}^{3}:7Fn+23−Fn3−Fn+13 in terms of xxx and yyy.We get:7(x+y)3−x3−y3: 7(x+y)^{3}-x^{3}-y^{3}:7(x+y)3−x3−y3.Simplifying that,we get:6x3+6y3+21xy(x+y).: 6x^{3}+6y^{3}+21xy(x+y).:6x3+6y3+21xy(x+y).Taking (x+y)(x+y)(x+y) common we get:(x+y)(6x2−6xy+6y2+21xy)=(x+y)(6x2+15xy+6y2)=(x+y)(x+2y)(6x+3y).: (x+y)(6x^{2}-6xy+6y^{2}+21xy) =(x+y)(6x^{2}+15xy+6y^{2}) =(x+y)(x+2y)(6x+3y).:(x+y)(6x2−6xy+6y2+21xy)=(x+y)(6x2+15xy+6y2)=(x+y)(x+2y)(6x+3y).But,(2y+x)=Fn+3.(2y+x)=F_{n+3}.(2y+x)=Fn+3.hence proved:):).
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@Krishna Ar @Siddharth G
Absolutely right. This is wht I tried, but the X,YX,YX,Y substitution, Mindblowing :D
@Krishna Ar – thanx!!
Perfect! I didn't go for x,y and faced problems in factorizing. PS: Small Typo at the end.
@Siddharth G – thanx!!
@Siddharth G – yes thanku!!
@Siddharth G – The substitution certainly helped made it easier to manipulate.
@Calvin Lin – Yes, however, can we do this with induction?
@Siddharth G – Possibly, but I don't see why that would work, nor do I see a way to start.
There is very little here to motivate a solution by induction. Knowing divisibility by Fn F_n Fn doesn't tell you anything about divisibility by Fn+1 F_{n+1} Fn+1.
How did you do?
Not well, 1b 5a, b 6b and half of the 8th were good. I left 1a half done. How was your paper?
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. :(
Same problem with wet 3a. BTW are you sure that 6a is 'Yes'?
@Siddharth G – Oh Nono, I didnt understand it properly but gave some stupid explanation and put Yes. WBU? How did you do that question?
@Krishna Ar – 6a. Forthe transformation, we needed ΔB=+2\Delta B=+2ΔB=+2 and ΔO=0\Delta O=0ΔO=0. However with the allowed changes, these conditions were contradicting each other.
@Siddharth G – Well, as I said...I wasnt even considering that question right :P. How was NTSE? (Range of marks..?)
@Krishna Ar – Terrible! Expecting 114/140; expected cutoff~123
@Siddharth G – Oh, I am sorry for bringing that up. How did you predict the cutoff so soon? (Institute?)...And, Could you tell me the number of seats for general cat. in Delhi.
@Krishna Ar – Not really, mostly from the marks of competitive peers. The no. of seats was 60 last year, however, I think they are going to increase it. FIITJEE predicts it to be near 119 due to this.
@Siddharth G – Oh I see. Be optimistic though :)
@Krishna Ar – Nah, just moving on. Are you giving RMO?
@Siddharth G – Yes I am. But NMTC has almost you know, shattered the vital mathematical force in me :3
@Krishna Ar – Oh, please. How was your NTSE?
@Siddharth G – Good. xD. It ough'to be that good lest you wouldn't qualify in TN. Expecting 130+, let's see.(the results)
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Let us take Fn=x and Fn+1=y⇒ Fn+2=x+y⇒ Fn+3=x+2y.Now,let us expand the given expression::7Fn+23−Fn3−Fn+13 in terms of x and y.We get:7(x+y)3−x3−y3.Simplifying that,we get:6x3+6y3+21xy(x+y).Taking (x+y) common we get:(x+y)(6x2−6xy+6y2+21xy)=(x+y)(6x2+15xy+6y2)=(x+y)(x+2y)(6x+3y).But,(2y+x)=Fn+3.hence proved:):).
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@Krishna Ar @Siddharth G
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Absolutely right. This is wht I tried, but the X,Y substitution, Mindblowing :D
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Perfect! I didn't go for x,y and faced problems in factorizing. PS: Small Typo at the end.
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There is very little here to motivate a solution by induction. Knowing divisibility by Fn doesn't tell you anything about divisibility by Fn+1.
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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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. :(
Same problem with wet 3a. BTW are you sure that 6a is 'Yes'?
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ΔB=+2 and ΔO=0. However with the allowed changes, these conditions were contradicting each other.
6a. Forthe transformation, we neededLog in to reply
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