2

Consider the Fibonacci sequence $\{a_{n}\}$ Use mathematical induction to prove that $a_{n+1}a_{n-1}=(a_{n})^{2}+(-1)^{n}$

So far, I have tested the base case $n=1$ which is true. I am stuck on the inductive step where I plug in $k=n+1$.

$a_{n+2}a_{n}=(a_{n+1})^{2}+(-1)^{n+1}$

$a_{n+2}a_{n}=(a_{n+1})^{2}-(-1)^{n}$

I am unsure what the next step to take is.

Daniel Fischer
  • 206,697
John Doe
  • 41
  • 3
    You need to start by testing it works for $n=1$ ... If I were you I would add that by editing your post ... that way people see what you have done and you will get more comments and advice. – Mandelbrot Sep 03 '18 at 08:33
  • For a recursive formula you need to give values such as $a_0$ and $a_1$ – Mandelbrot Sep 03 '18 at 08:35
  • Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Sep 03 '18 at 08:36
  • Hint: What happens if you apply the Fibonacci recurrence to $a_{n+2}$ and one of the $a_{n+1}$s? – Jyrki Lahtonen Sep 03 '18 at 08:54
  • Also, when dealing with the Fibonacci sequence it often happens that you need to apply the induction hypothesis to two previous values (because the recurrence refers to two earlier numbers). I'm not saying that's necessarily the case here, but it is possibility you should keep in mind. The price we then have to pay is that we need to check two lowest values as the base case. For otherwise the domino effect won't get started. – Jyrki Lahtonen Sep 03 '18 at 08:57
  • 1
    See https://math.stackexchange.com/a/1928409/589 – lhf Sep 03 '18 at 13:36
  • Possible duplicate of Show $F_{n+1} \cdot F_{n-1} = F_n^2 + (-1)^n$ for all $n \in \mathbb{N}$ – Arnaud D. Sep 03 '18 at 16:25

3 Answers3

2

Recall that: $a_1=1$,$a_2=1$,$a_3=2$ .

For $n=2$: $$ a_3a_1=a_2^2+(-1)^2=2 $$

We assume the hypothesis valid for $n=k$, such that: $$ a_{k+1}a_{k-1}=a_k^2+(-1)^k $$

Let's find the value of $a_{k+2}a_{k}$. We can use the Fibonacci recursion: $a_{k+2}=a_{k+1}+a_k$. Therefore: $$ a_{k+2}a_{k}=(a_{k+1}+a_k)a_k=a_{k+1}a_{k}+a_k^2 $$ From the hypothesis: $$ a_{k+1}a_{k-1}=a_k^2+(-1)^k \Rightarrow a_k^2 = a_{k+1}a_{k-1}-(-1)^k $$

Thus: $$ a_{k+2}a_{k}=(a_{k+1}+a_k)a_k=a_{k+1}a_{k}+a_{k+1}a_{k-1}-(-1)^k $$ $$ a_{k+2}a_{k}=a_{k+1}a_{k}+a_{k+1}a_{k-1}+(-1)^{k+1}=a_{k+1}(a_{k}+a_{k-1})+(-1)^{k+1} $$ Apply Fibonacci recursion again to find: $$ a_{k+2}a_{k}=a_{k+1}^2+(-1)^{k+1} $$

Mefitico
  • 1,845
1

For $n=1$, $~~a_{2}a_{0}-a_1^2=-1$, so the statement is true for $n=1$. Let, the statement $P(n):$ $ a_{n+1}a_{n-1}-a_n^2=(-1)^n$ is true for $n=m$. We need to show that $P(m+1)$ is true. $$a_{m+2}a_{m}-a_{m+1}^2=(-1)^{m+1} $$

Using recurrence relation on $a_{m+2}=a_{m+1}+a_m$ and $a_m=a_{m+1}-a_{m-1}$, $$\require{cancel}{a_{m+2}a_m-a_{m+1}^2=(a_{m+1}+a_{m})(a_{m+1}-a_{m-1})-a_{m+1}^2\\ = \cancel{a_{m+1}^2}+a_ma_{m+1}-a_{m+1}a_{m-1}-a_ma_{m-1}-\cancel{a_{m+1}^2}\\=a_m(a_{m+1}-a_{m-1})-a_{m+1}a_{m-1}= a_m^2-a_{m+1}a_{m-1}=-(a_{m+1}a_{m-1}-a_m^2)=(-1)^{m+1} }$$ as by inductive argument $a_{m+1}a_{m-1}-a_m^2=(-1)^m$. Hence done!

tarit goswami
  • 3,009
  • 1
  • 12
  • 27
1

The base hypothesis is obviously true: $0\cdot1-1^2=(-1)^1$.

Now the given formula hints to establish $$F_{n+1}F_{n-1}-F_n^2=-(F_{n}F_{n-2}-F_{n-1}^2).$$

If we expand $F_{n+1}$ we get $$F_{n}F_{n-1}+F_{n-1}^2-F_n^2=-F_{n}F_{n-2}+F_{n-1}^2.$$

Then regrouping the two terms with $F_n$, $$F_{n}(F_{n-1}-F_n)+F_{n-1}^2=-F_{n}F_{n-2}+F_{n-1}^2$$

and we are done.