Proof by Induction of Sum of Squares of Fibonacci using Difference Opperators
$begingroup$
Consider the sequence of Fibonacci numbers ${F_n}_{ngeq0}$ where $F_0=0,F_1=1$ and $F_n=F_{n-1}+F_{n-2}$ for $ngeq2$. It is proved that
begin{equation}sum_{i=0}^nF_i^2=F_nF_{n+1}.end{equation}
Suppose we generalize the Fibonacci numbers so that we have a sequence ${A_n}_{ngeq0}$ such that $A_0=0, A_1=1$ and $A_{n}=aA_{n-1}+bA_{n-2}$ for $ngeq2$. We will assume that $b ne 0$. We can show that
$$b^nA_0^2+b^{n-1}A_1^2+cdots+bA_{n-1}^2+A_n^2=frac{A_nA_{n+1}}{a}.$$
This formula can be proved by induction. I am not very good at proofs nor am I good at induction. If I could get some help that would be great.
sequences-and-series summation recurrence-relations fibonacci-numbers lucas-numbers
asked Dec 13 '18 at 19:16
User100290392039User100290392039
435
$endgroup$
add a comment |
$begingroup$
Consider the sequence of Fibonacci numbers ${F_n}_{ngeq0}$ where $F_0=0,F_1=1$ and $F_n=F_{n-1}+F_{n-2}$ for $ngeq2$. It is proved that
begin{equation}sum_{i=0}^nF_i^2=F_nF_{n+1}.end{equation}
Suppose we generalize the Fibonacci numbers so that we have a sequence ${A_n}_{ngeq0}$ such that $A_0=0, A_1=1$ and $A_{n}=aA_{n-1}+bA_{n-2}$ for $ngeq2$. We will assume that $b ne 0$. We can show that
$$b^nA_0^2+b^{n-1}A_1^2+cdots+bA_{n-1}^2+A_n^2=frac{A_nA_{n+1}}{a}.$$
This formula can be proved by induction. I am not very good at proofs nor am I good at induction. If I could get some help that would be great.
sequences-and-series summation recurrence-relations fibonacci-numbers lucas-numbers
asked Dec 13 '18 at 19:16
User100290392039User100290392039
435
$endgroup$
$begingroup$
What should be next to $sum_i^n$?
$endgroup$
– J.G.
Dec 13 '18 at 19:25
$begingroup$
I edited the problem accordingly, sorry for the mistake!
$endgroup$
– User100290392039
Dec 13 '18 at 19:28
$begingroup$
is little a also defined to be non-zero?
$endgroup$
– T. Fo
Dec 13 '18 at 19:38
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I am not too sure. I have taken this from The Fibonacci Numbers :Exposed math article by Kalman and Mena.
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– User100290392039
Dec 13 '18 at 19:46
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@User100290392039 You could use the explicit formula for the $n^{th}$ term of the Fibonacci sequence
$endgroup$
– Shubham Johri
Dec 13 '18 at 20:03
add a comment |
$begingroup$
Consider the sequence of Fibonacci numbers ${F_n}_{ngeq0}$ where $F_0=0,F_1=1$ and $F_n=F_{n-1}+F_{n-2}$ for $ngeq2$. It is proved that
begin{equation}sum_{i=0}^nF_i^2=F_nF_{n+1}.end{equation}
Suppose we generalize the Fibonacci numbers so that we have a sequence ${A_n}_{ngeq0}$ such that $A_0=0, A_1=1$ and $A_{n}=aA_{n-1}+bA_{n-2}$ for $ngeq2$. We will assume that $b ne 0$. We can show that
$$b^nA_0^2+b^{n-1}A_1^2+cdots+bA_{n-1}^2+A_n^2=frac{A_nA_{n+1}}{a}.$$
This formula can be proved by induction. I am not very good at proofs nor am I good at induction. If I could get some help that would be great.
sequences-and-series summation recurrence-relations fibonacci-numbers lucas-numbers
asked Dec 13 '18 at 19:16
User100290392039User100290392039
435
$endgroup$
Consider the sequence of Fibonacci numbers ${F_n}_{ngeq0}$ where $F_0=0,F_1=1$ and $F_n=F_{n-1}+F_{n-2}$ for $ngeq2$. It is proved that
begin{equation}sum_{i=0}^nF_i^2=F_nF_{n+1}.end{equation}
Suppose we generalize the Fibonacci numbers so that we have a sequence ${A_n}_{ngeq0}$ such that $A_0=0, A_1=1$ and $A_{n}=aA_{n-1}+bA_{n-2}$ for $ngeq2$. We will assume that $b ne 0$. We can show that
$$b^nA_0^2+b^{n-1}A_1^2+cdots+bA_{n-1}^2+A_n^2=frac{A_nA_{n+1}}{a}.$$
This formula can be proved by induction. I am not very good at proofs nor am I good at induction. If I could get some help that would be great.
sequences-and-series summation recurrence-relations fibonacci-numbers lucas-numbers
sequences-and-series summation recurrence-relations fibonacci-numbers lucas-numbers
asked Dec 13 '18 at 19:16
User100290392039User100290392039
435
asked Dec 13 '18 at 19:16
User100290392039User100290392039
435
edited Dec 13 '18 at 22:20
user614671
asked Dec 13 '18 at 19:16
User100290392039User100290392039
435
asked Dec 13 '18 at 19:16
User100290392039User100290392039
435
asked Dec 13 '18 at 19:16
User100290392039User100290392039
435
435
$begingroup$
What should be next to $sum_i^n$?
$endgroup$
– J.G.
Dec 13 '18 at 19:25
$begingroup$
I edited the problem accordingly, sorry for the mistake!
$endgroup$
– User100290392039
Dec 13 '18 at 19:28
$begingroup$
is little a also defined to be non-zero?
$endgroup$
– T. Fo
Dec 13 '18 at 19:38
$begingroup$
I am not too sure. I have taken this from The Fibonacci Numbers :Exposed math article by Kalman and Mena.
$endgroup$
– User100290392039
Dec 13 '18 at 19:46
$begingroup$
@User100290392039 You could use the explicit formula for the $n^{th}$ term of the Fibonacci sequence
$endgroup$
– Shubham Johri
Dec 13 '18 at 20:03
add a comment |
$begingroup$
What should be next to $sum_i^n$?
$endgroup$
– J.G.
Dec 13 '18 at 19:25
$begingroup$
I edited the problem accordingly, sorry for the mistake!
$endgroup$
– User100290392039
Dec 13 '18 at 19:28
$begingroup$
is little a also defined to be non-zero?
$endgroup$
– T. Fo
Dec 13 '18 at 19:38
$begingroup$
I am not too sure. I have taken this from The Fibonacci Numbers :Exposed math article by Kalman and Mena.
$endgroup$
– User100290392039
Dec 13 '18 at 19:46
$begingroup$
@User100290392039 You could use the explicit formula for the $n^{th}$ term of the Fibonacci sequence
$endgroup$
– Shubham Johri
Dec 13 '18 at 20:03
$begingroup$
What should be next to $sum_i^n$?
$endgroup$
– J.G.
Dec 13 '18 at 19:25
$begingroup$
What should be next to $sum_i^n$?
$endgroup$
– J.G.
Dec 13 '18 at 19:25
$begingroup$
I edited the problem accordingly, sorry for the mistake!
$endgroup$
– User100290392039
Dec 13 '18 at 19:28
$begingroup$
I edited the problem accordingly, sorry for the mistake!
$endgroup$
– User100290392039
Dec 13 '18 at 19:28
$begingroup$
is little a also defined to be non-zero?
$endgroup$
– T. Fo
Dec 13 '18 at 19:38
$begingroup$
is little a also defined to be non-zero?
$endgroup$
– T. Fo
Dec 13 '18 at 19:38
$begingroup$
I am not too sure. I have taken this from The Fibonacci Numbers :Exposed math article by Kalman and Mena.
$endgroup$
– User100290392039
Dec 13 '18 at 19:46
$begingroup$
I am not too sure. I have taken this from The Fibonacci Numbers :Exposed math article by Kalman and Mena.
$endgroup$
– User100290392039
Dec 13 '18 at 19:46
$begingroup$
@User100290392039 You could use the explicit formula for the $n^{th}$ term of the Fibonacci sequence
$endgroup$
– Shubham Johri
Dec 13 '18 at 20:03
$begingroup$
@User100290392039 You could use the explicit formula for the $n^{th}$ term of the Fibonacci sequence
$endgroup$
– Shubham Johri
Dec 13 '18 at 20:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let ${A_n}_{ngeq 0}$ be s.t. $A_0=0$, $A_1=1$, and $A_{n+2}=a A_{n+1}+b A_n$ for all non-negative integer $n$. We'll prove by induction on $n$ that
$$(1) asum_{k=0}^n b^{n-k}A_k^2=A_nA_{n+1}.$$
I do not see why we need the restriction $bneq 0$. If we treat $0^0$ as $1$, then (1) holds even in this case.
The claim is trivial for $n=0$ (since we have $0=0$). For $n=1$, we have $$a(bA_0+A_1)=a(0+1)=a=1cdot a=A_1A_2.$$
Suppose (1) holds for $n$. Then,
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=absum_{k=0}^nb^{n-k}A_k^2+aA_{n+1}^2.$$
By induction $$absum_{k=0}^nb^{n-k}A_k^2=bleft(asum_{k=0}^nb^{n-k}A_k^2right)=bleft(A_nA_{n+1}right).$$ That is,
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=b(A_nA_{n+1})+aA_{n+1}^2=A_{n+1}(aA_{n+1}+bA_n).$$
Since $A_{n+2}=aA_{n+1}+bA_n$, we obtain
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=A_{n+1}A_{n+2}$$
and the induction is complete.
$endgroup$
$begingroup$
How did you get $$(1) asum_{k=0}^n b^{n-k}A_k^2=A_nA_{n+1}.$$
$endgroup$
– User100290392039
Dec 14 '18 at 4:04
$begingroup$
@User100290392039 That is the induction hypothesis.
$endgroup$
– user614671
Dec 15 '18 at 18:18
add a comment |
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1 Answer
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$begingroup$
Let ${A_n}_{ngeq 0}$ be s.t. $A_0=0$, $A_1=1$, and $A_{n+2}=a A_{n+1}+b A_n$ for all non-negative integer $n$. We'll prove by induction on $n$ that
$$(1) asum_{k=0}^n b^{n-k}A_k^2=A_nA_{n+1}.$$
I do not see why we need the restriction $bneq 0$. If we treat $0^0$ as $1$, then (1) holds even in this case.
The claim is trivial for $n=0$ (since we have $0=0$). For $n=1$, we have $$a(bA_0+A_1)=a(0+1)=a=1cdot a=A_1A_2.$$
Suppose (1) holds for $n$. Then,
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=absum_{k=0}^nb^{n-k}A_k^2+aA_{n+1}^2.$$
By induction $$absum_{k=0}^nb^{n-k}A_k^2=bleft(asum_{k=0}^nb^{n-k}A_k^2right)=bleft(A_nA_{n+1}right).$$ That is,
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=b(A_nA_{n+1})+aA_{n+1}^2=A_{n+1}(aA_{n+1}+bA_n).$$
Since $A_{n+2}=aA_{n+1}+bA_n$, we obtain
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=A_{n+1}A_{n+2}$$
and the induction is complete.
$endgroup$
$begingroup$
How did you get $$(1) asum_{k=0}^n b^{n-k}A_k^2=A_nA_{n+1}.$$
$endgroup$
– User100290392039
Dec 14 '18 at 4:04
$begingroup$
@User100290392039 That is the induction hypothesis.
$endgroup$
– user614671
Dec 15 '18 at 18:18
add a comment |
$begingroup$
Let ${A_n}_{ngeq 0}$ be s.t. $A_0=0$, $A_1=1$, and $A_{n+2}=a A_{n+1}+b A_n$ for all non-negative integer $n$. We'll prove by induction on $n$ that
$$(1) asum_{k=0}^n b^{n-k}A_k^2=A_nA_{n+1}.$$
I do not see why we need the restriction $bneq 0$. If we treat $0^0$ as $1$, then (1) holds even in this case.
The claim is trivial for $n=0$ (since we have $0=0$). For $n=1$, we have $$a(bA_0+A_1)=a(0+1)=a=1cdot a=A_1A_2.$$
Suppose (1) holds for $n$. Then,
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=absum_{k=0}^nb^{n-k}A_k^2+aA_{n+1}^2.$$
By induction $$absum_{k=0}^nb^{n-k}A_k^2=bleft(asum_{k=0}^nb^{n-k}A_k^2right)=bleft(A_nA_{n+1}right).$$ That is,
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=b(A_nA_{n+1})+aA_{n+1}^2=A_{n+1}(aA_{n+1}+bA_n).$$
Since $A_{n+2}=aA_{n+1}+bA_n$, we obtain
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=A_{n+1}A_{n+2}$$
and the induction is complete.
$endgroup$
$begingroup$
How did you get $$(1) asum_{k=0}^n b^{n-k}A_k^2=A_nA_{n+1}.$$
$endgroup$
– User100290392039
Dec 14 '18 at 4:04
$begingroup$
@User100290392039 That is the induction hypothesis.
$endgroup$
– user614671
Dec 15 '18 at 18:18
add a comment |
$begingroup$
Let ${A_n}_{ngeq 0}$ be s.t. $A_0=0$, $A_1=1$, and $A_{n+2}=a A_{n+1}+b A_n$ for all non-negative integer $n$. We'll prove by induction on $n$ that
$$(1) asum_{k=0}^n b^{n-k}A_k^2=A_nA_{n+1}.$$
I do not see why we need the restriction $bneq 0$. If we treat $0^0$ as $1$, then (1) holds even in this case.
The claim is trivial for $n=0$ (since we have $0=0$). For $n=1$, we have $$a(bA_0+A_1)=a(0+1)=a=1cdot a=A_1A_2.$$
Suppose (1) holds for $n$. Then,
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=absum_{k=0}^nb^{n-k}A_k^2+aA_{n+1}^2.$$
By induction $$absum_{k=0}^nb^{n-k}A_k^2=bleft(asum_{k=0}^nb^{n-k}A_k^2right)=bleft(A_nA_{n+1}right).$$ That is,
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=b(A_nA_{n+1})+aA_{n+1}^2=A_{n+1}(aA_{n+1}+bA_n).$$
Since $A_{n+2}=aA_{n+1}+bA_n$, we obtain
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=A_{n+1}A_{n+2}$$
and the induction is complete.
$endgroup$
Let ${A_n}_{ngeq 0}$ be s.t. $A_0=0$, $A_1=1$, and $A_{n+2}=a A_{n+1}+b A_n$ for all non-negative integer $n$. We'll prove by induction on $n$ that
$$(1) asum_{k=0}^n b^{n-k}A_k^2=A_nA_{n+1}.$$
I do not see why we need the restriction $bneq 0$. If we treat $0^0$ as $1$, then (1) holds even in this case.
The claim is trivial for $n=0$ (since we have $0=0$). For $n=1$, we have $$a(bA_0+A_1)=a(0+1)=a=1cdot a=A_1A_2.$$
Suppose (1) holds for $n$. Then,
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=absum_{k=0}^nb^{n-k}A_k^2+aA_{n+1}^2.$$
By induction $$absum_{k=0}^nb^{n-k}A_k^2=bleft(asum_{k=0}^nb^{n-k}A_k^2right)=bleft(A_nA_{n+1}right).$$ That is,
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=b(A_nA_{n+1})+aA_{n+1}^2=A_{n+1}(aA_{n+1}+bA_n).$$
Since $A_{n+2}=aA_{n+1}+bA_n$, we obtain
$$asum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=A_{n+1}A_{n+2}$$
and the induction is complete.
answered Dec 13 '18 at 22:20
user614671
$begingroup$
How did you get $$(1) asum_{k=0}^n b^{n-k}A_k^2=A_nA_{n+1}.$$
$endgroup$
– User100290392039
Dec 14 '18 at 4:04
$begingroup$
@User100290392039 That is the induction hypothesis.
$endgroup$
– user614671
Dec 15 '18 at 18:18
add a comment |
$begingroup$
How did you get $$(1) asum_{k=0}^n b^{n-k}A_k^2=A_nA_{n+1}.$$
$endgroup$
– User100290392039
Dec 14 '18 at 4:04
$begingroup$
@User100290392039 That is the induction hypothesis.
$endgroup$
– user614671
Dec 15 '18 at 18:18
$begingroup$
How did you get $$(1) asum_{k=0}^n b^{n-k}A_k^2=A_nA_{n+1}.$$
$endgroup$
– User100290392039
Dec 14 '18 at 4:04
$begingroup$
How did you get $$(1) asum_{k=0}^n b^{n-k}A_k^2=A_nA_{n+1}.$$
$endgroup$
– User100290392039
Dec 14 '18 at 4:04
$begingroup$
@User100290392039 That is the induction hypothesis.
$endgroup$
– user614671
Dec 15 '18 at 18:18
$begingroup$
@User100290392039 That is the induction hypothesis.
$endgroup$
– user614671
Dec 15 '18 at 18:18
add a comment |
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$begingroup$
What should be next to $sum_i^n$?
$endgroup$
– J.G.
Dec 13 '18 at 19:25
$begingroup$
I edited the problem accordingly, sorry for the mistake!
$endgroup$
– User100290392039
Dec 13 '18 at 19:28
$begingroup$
is little a also defined to be non-zero?
$endgroup$
– T. Fo
Dec 13 '18 at 19:38
$begingroup$
I am not too sure. I have taken this from The Fibonacci Numbers :Exposed math article by Kalman and Mena.
$endgroup$
– User100290392039
Dec 13 '18 at 19:46
$begingroup$
@User100290392039 You could use the explicit formula for the $n^{th}$ term of the Fibonacci sequence
$endgroup$
– Shubham Johri
Dec 13 '18 at 20:03