Strong induction. Fibonacci numbers
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Using Strong Induction, prove that the (n + 3)-rd Fibonacci number can be computed as 1 plus the sum of the first n + 1 Fibonacci numbers (remember n includes 0). so it has to be proven by Strong Induction with the Inductive Hypothesis applied twice . . . twice because of the nature of the two-term recurrence.
Here I'm confused. I know how to use simple induction.
But I have to use strong induction.
The inductive hypothesis will be like
For arbitrary k ∈ N, ∀j ∈ N, 1 ≤ j ≤ k, S(j)
and the inductive step should be like
(∀j ∈ N, 1 ≤ j ≤ k, S(j)) → S(k + 1) then???
Examples can be n =5, sum of first 5 fibonacci numbers are 7 + 1 = 8. There is a fibonacci number 8 which is (5+3).
(n+3)rd fibonacci number = F(n + 2)because n = F(n-1)
discrete-mathematics induction fibonacci-numbers
edited Jan 17 at 23:09
Alexander Gruber♦
20k25102172
asked Oct 30 '18 at 1:33
dis mathdis math
13
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add a comment |
$begingroup$
Using Strong Induction, prove that the (n + 3)-rd Fibonacci number can be computed as 1 plus the sum of the first n + 1 Fibonacci numbers (remember n includes 0). so it has to be proven by Strong Induction with the Inductive Hypothesis applied twice . . . twice because of the nature of the two-term recurrence.
Here I'm confused. I know how to use simple induction.
But I have to use strong induction.
The inductive hypothesis will be like
For arbitrary k ∈ N, ∀j ∈ N, 1 ≤ j ≤ k, S(j)
and the inductive step should be like
(∀j ∈ N, 1 ≤ j ≤ k, S(j)) → S(k + 1) then???
Examples can be n =5, sum of first 5 fibonacci numbers are 7 + 1 = 8. There is a fibonacci number 8 which is (5+3).
(n+3)rd fibonacci number = F(n + 2)because n = F(n-1)
discrete-mathematics induction fibonacci-numbers
edited Jan 17 at 23:09
Alexander Gruber♦
20k25102172
asked Oct 30 '18 at 1:33
dis mathdis math
13
$endgroup$
2
$begingroup$
The specific result you are referring to can be proven using regular induction. You could of course phrase it using strong induction, but you probably won't use anything more than $S(k)$ to prove $S(k+1)$. As for a hint, remember that $f_{n}+f_{n+1}=f_{n+2}$.
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– JMoravitz
Oct 30 '18 at 1:43
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This tutorial explains how to typeset mathematics on this site.
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– N. F. Taussig
Oct 30 '18 at 10:48
add a comment |
$begingroup$
Using Strong Induction, prove that the (n + 3)-rd Fibonacci number can be computed as 1 plus the sum of the first n + 1 Fibonacci numbers (remember n includes 0). so it has to be proven by Strong Induction with the Inductive Hypothesis applied twice . . . twice because of the nature of the two-term recurrence.
Here I'm confused. I know how to use simple induction.
But I have to use strong induction.
The inductive hypothesis will be like
For arbitrary k ∈ N, ∀j ∈ N, 1 ≤ j ≤ k, S(j)
and the inductive step should be like
(∀j ∈ N, 1 ≤ j ≤ k, S(j)) → S(k + 1) then???
Examples can be n =5, sum of first 5 fibonacci numbers are 7 + 1 = 8. There is a fibonacci number 8 which is (5+3).
(n+3)rd fibonacci number = F(n + 2)because n = F(n-1)
discrete-mathematics induction fibonacci-numbers
edited Jan 17 at 23:09
Alexander Gruber♦
20k25102172
asked Oct 30 '18 at 1:33
dis mathdis math
13
$endgroup$
Using Strong Induction, prove that the (n + 3)-rd Fibonacci number can be computed as 1 plus the sum of the first n + 1 Fibonacci numbers (remember n includes 0). so it has to be proven by Strong Induction with the Inductive Hypothesis applied twice . . . twice because of the nature of the two-term recurrence.
Here I'm confused. I know how to use simple induction.
But I have to use strong induction.
The inductive hypothesis will be like
For arbitrary k ∈ N, ∀j ∈ N, 1 ≤ j ≤ k, S(j)
and the inductive step should be like
(∀j ∈ N, 1 ≤ j ≤ k, S(j)) → S(k + 1) then???
Examples can be n =5, sum of first 5 fibonacci numbers are 7 + 1 = 8. There is a fibonacci number 8 which is (5+3).
(n+3)rd fibonacci number = F(n + 2)because n = F(n-1)
discrete-mathematics induction fibonacci-numbers
discrete-mathematics induction fibonacci-numbers
edited Jan 17 at 23:09
Alexander Gruber♦
20k25102172
asked Oct 30 '18 at 1:33
dis mathdis math
13
edited Jan 17 at 23:09
Alexander Gruber♦
20k25102172
asked Oct 30 '18 at 1:33
dis mathdis math
13
edited Jan 17 at 23:09
Alexander Gruber♦
20k25102172
edited Jan 17 at 23:09
Alexander Gruber♦
20k25102172
edited Jan 17 at 23:09
Alexander Gruber♦
20k25102172
20k25102172
asked Oct 30 '18 at 1:33
dis mathdis math
13
asked Oct 30 '18 at 1:33
dis mathdis math
13
asked Oct 30 '18 at 1:33
dis mathdis math
13
13
2
$begingroup$
The specific result you are referring to can be proven using regular induction. You could of course phrase it using strong induction, but you probably won't use anything more than $S(k)$ to prove $S(k+1)$. As for a hint, remember that $f_{n}+f_{n+1}=f_{n+2}$.
$endgroup$
– JMoravitz
Oct 30 '18 at 1:43
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Oct 30 '18 at 10:48
add a comment |
2
$begingroup$
The specific result you are referring to can be proven using regular induction. You could of course phrase it using strong induction, but you probably won't use anything more than $S(k)$ to prove $S(k+1)$. As for a hint, remember that $f_{n}+f_{n+1}=f_{n+2}$.
$endgroup$
– JMoravitz
Oct 30 '18 at 1:43
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Oct 30 '18 at 10:48
2
2
$begingroup$
The specific result you are referring to can be proven using regular induction. You could of course phrase it using strong induction, but you probably won't use anything more than $S(k)$ to prove $S(k+1)$. As for a hint, remember that $f_{n}+f_{n+1}=f_{n+2}$.
$endgroup$
– JMoravitz
Oct 30 '18 at 1:43
$begingroup$
The specific result you are referring to can be proven using regular induction. You could of course phrase it using strong induction, but you probably won't use anything more than $S(k)$ to prove $S(k+1)$. As for a hint, remember that $f_{n}+f_{n+1}=f_{n+2}$.
$endgroup$
– JMoravitz
Oct 30 '18 at 1:43
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Oct 30 '18 at 10:48
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Oct 30 '18 at 10:48
add a comment |
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The specific result you are referring to can be proven using regular induction. You could of course phrase it using strong induction, but you probably won't use anything more than $S(k)$ to prove $S(k+1)$. As for a hint, remember that $f_{n}+f_{n+1}=f_{n+2}$.
$endgroup$
– JMoravitz
Oct 30 '18 at 1:43
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Oct 30 '18 at 10:48