Prove for all $nin mathbbN$ that $sum_i=0^n icdot F_2i = (n+1)F_2n + 1 - F_2n + 2$.Showing that an equation holds true with a Fibonacci sequence: $F_n+m = F_n-1F_m + F_n F_m+1$Proof by Induction: Alternating Sum of Fibonacci NumbersFibonacci sequencesQuestion regarding the Fibonacci sequenceFibonacci numeration systemHelp with how to prepare the inductive step of a strong induction exercise.Fibonacci proof question $sum_i=1^nF_i = F_n+2 - 1$Inductively proving a fibonacci numbers statementSeries with inverse Fibonacci productsStrong Inductive proof for inequality using Fibonacci sequence
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Prove for all $nin mathbbN$ that $sum_i=0^n icdot F_2i = (n+1)F_2n + 1 - F_2n + 2$.
Showing that an equation holds true with a Fibonacci sequence: $F_n+m = F_n-1F_m + F_n F_m+1$Proof by Induction: Alternating Sum of Fibonacci NumbersFibonacci sequencesQuestion regarding the Fibonacci sequenceFibonacci numeration systemHelp with how to prepare the inductive step of a strong induction exercise.Fibonacci proof question $sum_i=1^nF_i = F_n+2 - 1$Inductively proving a fibonacci numbers statementSeries with inverse Fibonacci productsStrong Inductive proof for inequality using Fibonacci sequence
$begingroup$
$F_n$ denotes the Fibonacci sequence where $n$ is the term of the Fibonacci number in the sequence. ($F_0=0$, $F_1=1$, $F_2=1$, $F_3=2$, $F_4=3$, ... $F_n=F_n-1 + F_n-2$)
I want to prove this using strong induction and have formulated base cases for when $n=0$ and when $n=1$.
My induction hypothesis is that for all $kin mathbbN$, $kgeq 1$ we can assume that for $k-1$, $sum_i=0^k-1 icdot F_2i = (k)F_2k - 1 - F_2k$ and that for $k$, $sum_i=0^k icdot F_2i = (k + 1)F_2k + 1 - F_2k + 2$.
For my induction step I want to show that for $k+1$, $sum_i=0^k+1 icdot F_2i = (k+2)F_2k + 3 - F_2k + 4$. I've been stuck on the equation below for the past couple of hours trying to figure out how I can derive the equation for $k + 1$.
beginalign*
sum_i=0^k+1 icdot F_2i &= (k + 1)cdot F_2k + 2 + sum_i=0^k icdot F_2i\
&= (k + 1)cdot F_2k + 2 + (k + 1)F_2k + 1 - F_2k + 2 + sum_i=0^k - 1 icdot F_2i\
&= (k + 1)cdot F_2k + 2 + (k + 1)F_2k + 1 - F_2k + 2 + (k)F_2k - 1 - F_2k\
endalign*
Any help is appreciated.
discrete-mathematics induction fibonacci-numbers
asked Mar 17 at 22:45
ivyleaf57ivyleaf57
514
$endgroup$
add a comment |
$begingroup$
$F_n$ denotes the Fibonacci sequence where $n$ is the term of the Fibonacci number in the sequence. ($F_0=0$, $F_1=1$, $F_2=1$, $F_3=2$, $F_4=3$, ... $F_n=F_n-1 + F_n-2$)
I want to prove this using strong induction and have formulated base cases for when $n=0$ and when $n=1$.
My induction hypothesis is that for all $kin mathbbN$, $kgeq 1$ we can assume that for $k-1$, $sum_i=0^k-1 icdot F_2i = (k)F_2k - 1 - F_2k$ and that for $k$, $sum_i=0^k icdot F_2i = (k + 1)F_2k + 1 - F_2k + 2$.
For my induction step I want to show that for $k+1$, $sum_i=0^k+1 icdot F_2i = (k+2)F_2k + 3 - F_2k + 4$. I've been stuck on the equation below for the past couple of hours trying to figure out how I can derive the equation for $k + 1$.
beginalign*
sum_i=0^k+1 icdot F_2i &= (k + 1)cdot F_2k + 2 + sum_i=0^k icdot F_2i\
&= (k + 1)cdot F_2k + 2 + (k + 1)F_2k + 1 - F_2k + 2 + sum_i=0^k - 1 icdot F_2i\
&= (k + 1)cdot F_2k + 2 + (k + 1)F_2k + 1 - F_2k + 2 + (k)F_2k - 1 - F_2k\
endalign*
Any help is appreciated.
discrete-mathematics induction fibonacci-numbers
asked Mar 17 at 22:45
ivyleaf57ivyleaf57
514
$endgroup$
add a comment |
$begingroup$
$F_n$ denotes the Fibonacci sequence where $n$ is the term of the Fibonacci number in the sequence. ($F_0=0$, $F_1=1$, $F_2=1$, $F_3=2$, $F_4=3$, ... $F_n=F_n-1 + F_n-2$)
I want to prove this using strong induction and have formulated base cases for when $n=0$ and when $n=1$.
My induction hypothesis is that for all $kin mathbbN$, $kgeq 1$ we can assume that for $k-1$, $sum_i=0^k-1 icdot F_2i = (k)F_2k - 1 - F_2k$ and that for $k$, $sum_i=0^k icdot F_2i = (k + 1)F_2k + 1 - F_2k + 2$.
For my induction step I want to show that for $k+1$, $sum_i=0^k+1 icdot F_2i = (k+2)F_2k + 3 - F_2k + 4$. I've been stuck on the equation below for the past couple of hours trying to figure out how I can derive the equation for $k + 1$.
beginalign*
sum_i=0^k+1 icdot F_2i &= (k + 1)cdot F_2k + 2 + sum_i=0^k icdot F_2i\
&= (k + 1)cdot F_2k + 2 + (k + 1)F_2k + 1 - F_2k + 2 + sum_i=0^k - 1 icdot F_2i\
&= (k + 1)cdot F_2k + 2 + (k + 1)F_2k + 1 - F_2k + 2 + (k)F_2k - 1 - F_2k\
endalign*
Any help is appreciated.
discrete-mathematics induction fibonacci-numbers
asked Mar 17 at 22:45
ivyleaf57ivyleaf57
514
$endgroup$
$F_n$ denotes the Fibonacci sequence where $n$ is the term of the Fibonacci number in the sequence. ($F_0=0$, $F_1=1$, $F_2=1$, $F_3=2$, $F_4=3$, ... $F_n=F_n-1 + F_n-2$)
I want to prove this using strong induction and have formulated base cases for when $n=0$ and when $n=1$.
My induction hypothesis is that for all $kin mathbbN$, $kgeq 1$ we can assume that for $k-1$, $sum_i=0^k-1 icdot F_2i = (k)F_2k - 1 - F_2k$ and that for $k$, $sum_i=0^k icdot F_2i = (k + 1)F_2k + 1 - F_2k + 2$.
For my induction step I want to show that for $k+1$, $sum_i=0^k+1 icdot F_2i = (k+2)F_2k + 3 - F_2k + 4$. I've been stuck on the equation below for the past couple of hours trying to figure out how I can derive the equation for $k + 1$.
beginalign*
sum_i=0^k+1 icdot F_2i &= (k + 1)cdot F_2k + 2 + sum_i=0^k icdot F_2i\
&= (k + 1)cdot F_2k + 2 + (k + 1)F_2k + 1 - F_2k + 2 + sum_i=0^k - 1 icdot F_2i\
&= (k + 1)cdot F_2k + 2 + (k + 1)F_2k + 1 - F_2k + 2 + (k)F_2k - 1 - F_2k\
endalign*
Any help is appreciated.
discrete-mathematics induction fibonacci-numbers
discrete-mathematics induction fibonacci-numbers
asked Mar 17 at 22:45
ivyleaf57ivyleaf57
514
asked Mar 17 at 22:45
ivyleaf57ivyleaf57
514
asked Mar 17 at 22:45
ivyleaf57ivyleaf57
514
asked Mar 17 at 22:45
ivyleaf57ivyleaf57
514
asked Mar 17 at 22:45
ivyleaf57ivyleaf57
514
514
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
beginalign
sum_i=0^k+1iF_2i&=(k+1)F_2k+2+(k+1)F_2k+1-F_2k+2\
&=kF_2k+2+(k+1)(F_2k+3-F_2k+2)\
&=(k+1)F_2k+3-F_2k+2\
&=(k+1)F_2k+3-(F_2k+4-F_2k+3)\
&=(k+2)F_2k+3-F_2k+4.
endalign
answered Mar 17 at 22:55
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Thanks, It seems like you were able to get the answer using ordinary induction. I was told that whenever you see Fibonacci numbers you should go straight to strong induction. Is there a way to know which you should start with? Or should you always start with ordinary induction and then move to strong induction if you need to assume more?
$endgroup$
– ivyleaf57
Mar 17 at 23:02
add a comment |
Your Answer
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1 Answer
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$begingroup$
beginalign
sum_i=0^k+1iF_2i&=(k+1)F_2k+2+(k+1)F_2k+1-F_2k+2\
&=kF_2k+2+(k+1)(F_2k+3-F_2k+2)\
&=(k+1)F_2k+3-F_2k+2\
&=(k+1)F_2k+3-(F_2k+4-F_2k+3)\
&=(k+2)F_2k+3-F_2k+4.
endalign
answered Mar 17 at 22:55
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Thanks, It seems like you were able to get the answer using ordinary induction. I was told that whenever you see Fibonacci numbers you should go straight to strong induction. Is there a way to know which you should start with? Or should you always start with ordinary induction and then move to strong induction if you need to assume more?
$endgroup$
– ivyleaf57
Mar 17 at 23:02
add a comment |
$begingroup$
beginalign
sum_i=0^k+1iF_2i&=(k+1)F_2k+2+(k+1)F_2k+1-F_2k+2\
&=kF_2k+2+(k+1)(F_2k+3-F_2k+2)\
&=(k+1)F_2k+3-F_2k+2\
&=(k+1)F_2k+3-(F_2k+4-F_2k+3)\
&=(k+2)F_2k+3-F_2k+4.
endalign
answered Mar 17 at 22:55
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Thanks, It seems like you were able to get the answer using ordinary induction. I was told that whenever you see Fibonacci numbers you should go straight to strong induction. Is there a way to know which you should start with? Or should you always start with ordinary induction and then move to strong induction if you need to assume more?
$endgroup$
– ivyleaf57
Mar 17 at 23:02
add a comment |
$begingroup$
beginalign
sum_i=0^k+1iF_2i&=(k+1)F_2k+2+(k+1)F_2k+1-F_2k+2\
&=kF_2k+2+(k+1)(F_2k+3-F_2k+2)\
&=(k+1)F_2k+3-F_2k+2\
&=(k+1)F_2k+3-(F_2k+4-F_2k+3)\
&=(k+2)F_2k+3-F_2k+4.
endalign
answered Mar 17 at 22:55
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
beginalign
sum_i=0^k+1iF_2i&=(k+1)F_2k+2+(k+1)F_2k+1-F_2k+2\
&=kF_2k+2+(k+1)(F_2k+3-F_2k+2)\
&=(k+1)F_2k+3-F_2k+2\
&=(k+1)F_2k+3-(F_2k+4-F_2k+3)\
&=(k+2)F_2k+3-F_2k+4.
endalign
answered Mar 17 at 22:55
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Mar 17 at 22:55
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Mar 17 at 22:55
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Mar 17 at 22:55
Lord Shark the UnknownLord Shark the Unknown
107k1162135
107k1162135
$begingroup$
Thanks, It seems like you were able to get the answer using ordinary induction. I was told that whenever you see Fibonacci numbers you should go straight to strong induction. Is there a way to know which you should start with? Or should you always start with ordinary induction and then move to strong induction if you need to assume more?
$endgroup$
– ivyleaf57
Mar 17 at 23:02
add a comment |
$begingroup$
Thanks, It seems like you were able to get the answer using ordinary induction. I was told that whenever you see Fibonacci numbers you should go straight to strong induction. Is there a way to know which you should start with? Or should you always start with ordinary induction and then move to strong induction if you need to assume more?
$endgroup$
– ivyleaf57
Mar 17 at 23:02
$begingroup$
Thanks, It seems like you were able to get the answer using ordinary induction. I was told that whenever you see Fibonacci numbers you should go straight to strong induction. Is there a way to know which you should start with? Or should you always start with ordinary induction and then move to strong induction if you need to assume more?
$endgroup$
– ivyleaf57
Mar 17 at 23:02
$begingroup$
Thanks, It seems like you were able to get the answer using ordinary induction. I was told that whenever you see Fibonacci numbers you should go straight to strong induction. Is there a way to know which you should start with? Or should you always start with ordinary induction and then move to strong induction if you need to assume more?
$endgroup$
– ivyleaf57
Mar 17 at 23:02
add a comment |
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