0

The Fibonacci sequence is defined as the sequence where $F_1 = 1,F_2= 1$ and $F_i=F_{i-1}+ F_{i-2}$. Use induction to prove the that for $n\ge 2$, $$F_1+F_2+ \ldots+F_{n-1}=F_{n+1}-1$$

gt6989b
  • 54,422
hhh
  • 1
  • please show your effort on what looks like a homework problem. Update your questions with your thoughts and where you got stuck and we will be happy to help you move forward. People here don't usually like to do your homework for you – gt6989b Dec 12 '16 at 01:36
  • Alrighty, so subbing in n+1 into our recursive definition we get: fn+1=f0+f-1, which is the farthest I've gotten – hhh Dec 12 '16 at 01:41
  • look at the hint in my answer and update the question with your progress – gt6989b Dec 12 '16 at 01:42

1 Answers1

0

HINT

All proofs by induction look like this:

Base Case For the smallest value of $n$ (what is it in your case?), prove your statement (rephrase it plugging in the needed value of $n$ and use $F_1=1=F_1$ to show this is true.

Inductive Step

Assume the statement is true for all $n$ up to some $N-1$. Now use this to prove your statement for $n=N$. (Hint: what does your statement say for $n=N-1$ -- you are assuming it is true; can you prove your statement from that?)

gt6989b
  • 54,422
  • For the base case, I would sub-in n=2, which gives me f1+f0... am I allowed to assume f0=zero? – hhh Dec 12 '16 at 01:45
  • You are trying to show $$F_1+ \ldots+F_{n-1}=F_{n+1}-1$$ What does this statement look like for $n=2$? – gt6989b Dec 12 '16 at 01:47
  • It would be F1+…+Fn−1=0 – hhh Dec 12 '16 at 01:48
  • be very specific - you know $n$, so your left-hand side cannot depend on $n$. Also why is the right hand side 0? – gt6989b Dec 12 '16 at 01:49
  • Which f2, is supposed to equal 1? So does this mean I've disproven the base case, thus disproving the whole statement or am I on the wrong train of thought? – hhh Dec 12 '16 at 01:49
  • My apologies, simply arithmetic mistake – hhh Dec 12 '16 at 01:49
  • At n=2, the statement looks like f1+...fn-1=f3-1 – hhh Dec 12 '16 at 01:51
  • Right, and then subbing that into your original you find that F3 gives you 2 and 2-1=1, so you've proven your base case for n=2 – hhh Dec 12 '16 at 01:53
  • For the conclusion you must show k+1=n is true – hhh Dec 12 '16 at 01:54
  • ok - try it and update it with your progress – gt6989b Dec 12 '16 at 02:01
  • Alright, well subbing k+1 for n we get f1+f2+...+fk=fk+2-1. Subbing for fk+3 in the recursive definition we get, fk+2= fk+1+fk – hhh Dec 12 '16 at 02:09
  • @hhh that last statement, $F_{k+2} = F_k + F_{K+1}$ is true by definition of the Fibonacci sequence, so this proves your statement... – gt6989b Dec 12 '16 at 03:47