Question

L(n) = { 1 if n = 1
L(n) = { 3 if n = 2
L(n) = { L(n − 1) + L(n − 2) if n > 2.
Let α =1 + sqrt5/ 2 and β =1 − sqrt5/ 2.
. Prove that L(n) = αn + βn for all n is in N. Use strong induction.
Proof. First, note that
L(1) = 1 = α + β,
and α^2 + β^2 = (α + 1) + (β + ?)
= α + β + 2
= ? = L(2)
Suppose as inductive hypothesis that L(i) = αi + βi for all i < k, for some k > 2. Then
L(k) = L(k − 1) + L(k −?)
=α^k − 1 + β^k − 1 + ( ? )
=α^k − 2(α + 1) + β^k − 2(β + ?)
=α^k − 2(α2) + β^k − 2( ? )
=α^k + ? , as required

Answer 1

visual to see if it makes more sense this way

Answer 2

well, what does L(2) equal to start with?

Answer 3

\[\frac{1+\sqrt5}{2}+\frac{1-\sqrt5}{2}+2\]
\[\frac{1+\sqrt5+1-\sqrt5}{2}+2\]
\[\frac{1+1}{2}+2\]
\[\frac{2}{2}+2\]
\[1+2\]...

Answer 4

its given that:

L(n) = L(n-1) + L(n-2) for n > 2 ; and when they replace n by k we get

L(k) = L(k-1) + L(k -[______] ) for k > 2
??

Answer 5

L(2) = 3

Answer 6

yes, and for this proof to work out the way it wants, the 3 = L(2)

Answer 7

Ok, I see how we got 3 = L(2) but now I am not seeing how L(k)=L(k − 1) + L(k - ____) the logical answer I see on this line is 2, but I think I am wrong.

Answer 8

your right, we are simply replacing n by k to make it more generalized is all

Answer 9

or should i say to make it more specific ...

n is a general element from the set: {1,2,3,4,....}

for some k in the set would make it more specific ... and since we know that for n=k=1 it works out; then if we can show it for the same format in k as for k+1 .. we have a starting place and can move forward

Answer 10

boxes 5,6,7 are pretty self explanatory ....

a+1 leads us to a b+1 setup

a^2 leads us to a b^2 setup

a^k leads us to a b^k setup

Answer 11

the 4th box i beleive should amount to:\[a^{k-2}+b^{k-2}\]

Answer 12

\[a^{k-1}+a^{k-2}\]
\[a^{k-2}(a^1+1)\]

Answer 13

ah ha!....I see, I see, thank you again, much appreciated, I like how you explain and not just answer it

Answer 14

lol, im just not smart enough to know the answer beforehand :)