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.
Is this better???
Step 1: For n = 1, we have
Hence the equation holds for n = 1
Step 2: Assume the equation is true for n = k (inductive hypothesis), we have:
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:
Since there is a such as,
Then proof it hold for . But the expression,
But,
Thus,
Thus, it hold true for .
Hence the equation holds true for n = k + 1
Hello, Natasha!
An inductive proof requires some world-class acrobatics . . .
I need to prove by induction that for the Fibonacci sequence:
. . . being a Fibonacci number
Verify -1)^2\quad\Rightarrow\quad 1\cdot2 - 1^2\:=\:1" alt="S(1):\;\;(F_1)\cdot(F_3) - (F_2)^2 \:=\-1)^2\quad\Rightarrow\quad 1\cdot2 - 1^2\:=\:1" /> . . . yes!
Assume is true: .
. . and we will try to prove F_{k+1})(F_{k+3}) - (F_{k+2})^2\:=\-1)^{k+2} " alt="S(k+1)\:=\F_{k+1})(F_{k+3}) - (F_{k+2})^2\:=\-1)^{k+2} " />
Since and
becomes:
Expand:
We have:
Factor:
. . . . . . . . . . .This is
An we have:
Multiply by
. . . ta-DAA! . . . This is