1. You are doing it correctly, but you need to manipulate that fibonacci expression. You can do that by looking at the non-inductive proof I gave. And following with the same reasoning.

2. Originally Posted by ThePerfectHacker
You are doing it correctly, but you need to manipulate that fibonacci expression. You can do that by looking at the non-inductive proof I gave. And following with the same reasoning.
I'm so rubbish at the induction proof. Could someone just show me what I need to do? It's due in for tomorrow please

3. Is this better???

Step 1: For n = 1, we have

$\displaystyle (F_1 \cdot F_{1+2}) - (F_{1+1})^2 = (-1)^{1+1} = 1$

Hence the equation holds for n = 1

Step 2: Assume the equation is true for n = k (inductive hypothesis), we have:

$\displaystyle (F_k \cdot F_{k+2}) - (F_{k+1})^2 = (-1)^{k+1}$

Step 3: We then have to use our inductive hypothesis to prove that the statement is also true for n = k + 1, i.e we have to prove:

$\displaystyle (F_{k+1} \cdot F_{k+3}) - (F_{k+2})^2 = (-1)^{k+2}$

Since there is a $\displaystyle k\geq 2$ such as,
$\displaystyle F(k)^2-F(k+1)F(k-1)=(-1)^{k-1}$
Then proof it hold for $\displaystyle k+1$. But the expression,
$\displaystyle F(k+1)^2-F(k+2)F(k)=(-1)(F(k)^2-F(k+1)F(k-1))$
But,
$\displaystyle F(k)^2-F(k+1)(k-1)=(-1)^{k-1}$
Thus,
$\displaystyle F(k+1)^2-F(k+1)F(k)=(-1)(-1)^{k-1}=(-1)^k$
Thus, it hold true for $\displaystyle k+1$.

$\displaystyle (F_{k+1} \cdot F_{k+3}) - (F_{k+2})^2 = (-1)^{k+2}$

Hence the equation holds true for n = k + 1

4. Originally Posted by Natasha1
Is this better???

Step 1: For n = 1, we have

$\displaystyle (F_1 \cdot F_{1+2}) - (F_{1+1})^2 = (-1)^{1+1} = 1$

Hence the equation holds for n = 1

Step 2: Assume the equation is true for n = k (inductive hypothesis), we have:

$\displaystyle (F_k \cdot F_{k+2}) - (F_{k+1})^2 = (-1)^{k+1}$

Step 3: We then have to use our inductive hypothesis to prove that the statement is also true for n = k + 1, i.e we have to prove:
$\displaystyle F_{k+1} F_{k+3} - (F_{k+2})^2 =$$\displaystyle F_{k+1} (F_{k+2}+F_{k+1}) - (F_{k+2})^2 \displaystyle =F_{k+1} F_{k+2}+(F_{k+1})^2 - (F_{k+2})^2. Now: \displaystyle F_{k+1}=F_{k+2}-F_{k}, so: \displaystyle F_{k+1} F_{k+3} - (F_{k+2})^2=$$\displaystyle (F_{k+2}-F_{k}) F_{k+2}+(F_{k+1})^2 - (F_{k+2})^2$

$\displaystyle =-F_{k} F_{k+2}+(F_{k+1})^2=-1((F_k \cdot F_{k+2}) - (F_{k+1})^2)=(-1)^{k+2}$
.

Which proves that if the statement is true for $\displaystyle k$ it is true for
$\displaystyle k+1$.

RonL

5. Hello, Natasha!

An inductive proof requires some world-class acrobatics . . .

I need to prove by induction that for the Fibonacci sequence:

$\displaystyle (F_n)\cdot(F_{n+2}) - (F_{n+1})^2 \;= \;(-1)^{n+1}$ . . . $\displaystyle F_n$ being a Fibonacci number

Verify $\displaystyle S(1):\;\;(F_1)\cdot(F_3) - (F_2)^2 \:=\-1)^2\quad\Rightarrow\quad 1\cdot2 - 1^2\:=\:1$ . . . yes!

Assume $\displaystyle S(k)$ is true: .$\displaystyle (F_k)(F_{k+2}) - (F_{k+1})^2\;=\;(-1)^{k+1}$

. . and we will try to prove $\displaystyle S(k+1)\:=\F_{k+1})(F_{k+3}) - (F_{k+2})^2\:=\-1)^{k+2}$

Since $\displaystyle F_k\,=\,F_{k+2} - F_{k+1}$ and $\displaystyle F_{k+2} \,=\,F_{k+3} - F_{k+1}$

$\displaystyle S(k)$ becomes: $\displaystyle (F_{k+2} - F_{k+1})(F_{k+3} - F_{k+1}) - (F_{k+1})^2 \;= \;(-1)^{k+1}$

Expand: $\displaystyle (F_{k+2})(F_{k+3}) - (F_{k+1})(F_{k+2}) - (F_{k+1})(F_{k+3}) + (F_{k+1})^2 -$$\displaystyle (F_{k+1})^2\;=\;(-1)^{k+1}$

We have: $\displaystyle (F_{k+2})(F_{k+3}) - (F_{k+1})(F_{k+2}) - (F_{k+1})(F_{k+3}) \;= \; (-1)^{k+1}$

Factor: $\displaystyle (F_{k+2})\cdot(\underbrace{F_{k+3} - F_{k+1}}) - (F_{k+1})(F_{k+3}) \;= \; (-1)^{k+1}$
. . . . . . . . . . .This is $\displaystyle F_{k+2}$

An we have: $\displaystyle (F_{k+2})^2 - (F_{k+1})(F_{k+3}) \;=\;(-1)^{k+1}$

Multiply by $\displaystyle -1:\;\;(F_{k+1})(F_{k+3}) - (F_{k+2})^2\;=\;(-1)^{k+2}$

. . . ta-DAA! . . . This is $\displaystyle S(k+1)\;!$

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