Prove using mathematical induction that $sum_i=1^n if_i = nf_n+2 - f_n+3 + 2 for all n geqq 1$Use the recursive definition of summation together with mathematical induction to prove a sequenceProve/disprove $n! = O(2^n)$ via mathematical inductionProve $ sum_k=0^n k4^k = frac 49((3n-1)4^n + 1) $ by inductionDiscrete Math Induction Proof Help With Questionproving the divisibility rule for 11 using inductionProve that Fibonacci numbers are greater than a certain valueFibonacci induction problem gone wrong!Proof by induction, Inductive step simplicifationProve by induction that $sum_i=1^n frac1sqrt i > sqrt n$ for all integers $n ge 2$Prove for all $nin mathbbN$ that $sum_i=0^n icdot F_2i = (n+1)F_2n + 1 - F_2n + 2$.
Prove that NP is closed under karp reduction?
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Prove using mathematical induction that $sum_i=1^n if_i = nf_n+2 - f_n+3 + 2 for all n geqq 1$
Use the recursive definition of summation together with mathematical induction to prove a sequenceProve/disprove $n! = O(2^n)$ via mathematical inductionProve $ sum_k=0^n k4^k = frac 49((3n-1)4^n + 1) $ by inductionDiscrete Math Induction Proof Help With Questionproving the divisibility rule for 11 using inductionProve that Fibonacci numbers are greater than a certain valueFibonacci induction problem gone wrong!Proof by induction, Inductive step simplicifationProve by induction that $sum_i=1^n frac1sqrt i > sqrt n$ for all integers $n ge 2$Prove for all $nin mathbbN$ that $sum_i=0^n icdot F_2i = (n+1)F_2n + 1 - F_2n + 2$.
$begingroup$
This question is for the Fibonacci sequence, where $mathit f $ is represented by the fibonacci function.
I have done my base step and elaborated my inductive hypothesis already, but I am having trouble with my inductive step.
$pmb Inductive Hypothesis:$ Let z $in$ $mathbb Z^+$ be given. Assume $sum_i=1^n if_i = nf_n+2 - f_n+3 + 2 $ is true for all n = z.
From what I believe, I must prove that $sum_i=1^z+1 if_i $ = $ (z+1)f_z+3 - f_z+4 + 2$.
I have reached the following:
beginalign
sum_i=1^z+1 if_i &= (sum_i=1^z if_i) + (z+1)f_z+1 \
& = zf_z+2 - f_z+3 + 2 + (z+1)f_z+1 pmb(by inductive hypothesis)
endalign
I do not know where to take it from this point on.
Any help would be greatly appreciated!
discrete-mathematics
asked Mar 29 at 15:07
IOStatIOStat
31
$endgroup$
add a comment |
$begingroup$
This question is for the Fibonacci sequence, where $mathit f $ is represented by the fibonacci function.
I have done my base step and elaborated my inductive hypothesis already, but I am having trouble with my inductive step.
$pmb Inductive Hypothesis:$ Let z $in$ $mathbb Z^+$ be given. Assume $sum_i=1^n if_i = nf_n+2 - f_n+3 + 2 $ is true for all n = z.
From what I believe, I must prove that $sum_i=1^z+1 if_i $ = $ (z+1)f_z+3 - f_z+4 + 2$.
I have reached the following:
beginalign
sum_i=1^z+1 if_i &= (sum_i=1^z if_i) + (z+1)f_z+1 \
& = zf_z+2 - f_z+3 + 2 + (z+1)f_z+1 pmb(by inductive hypothesis)
endalign
I do not know where to take it from this point on.
Any help would be greatly appreciated!
discrete-mathematics
asked Mar 29 at 15:07
IOStatIOStat
31
$endgroup$
$begingroup$
You need to show it is true for $n=0$, then you should use $f_n+f_n+1=f_n+2$.
$endgroup$
– Alec B-G
Mar 29 at 15:24
add a comment |
$begingroup$
This question is for the Fibonacci sequence, where $mathit f $ is represented by the fibonacci function.
I have done my base step and elaborated my inductive hypothesis already, but I am having trouble with my inductive step.
$pmb Inductive Hypothesis:$ Let z $in$ $mathbb Z^+$ be given. Assume $sum_i=1^n if_i = nf_n+2 - f_n+3 + 2 $ is true for all n = z.
From what I believe, I must prove that $sum_i=1^z+1 if_i $ = $ (z+1)f_z+3 - f_z+4 + 2$.
I have reached the following:
beginalign
sum_i=1^z+1 if_i &= (sum_i=1^z if_i) + (z+1)f_z+1 \
& = zf_z+2 - f_z+3 + 2 + (z+1)f_z+1 pmb(by inductive hypothesis)
endalign
I do not know where to take it from this point on.
Any help would be greatly appreciated!
discrete-mathematics
asked Mar 29 at 15:07
IOStatIOStat
31
$endgroup$
This question is for the Fibonacci sequence, where $mathit f $ is represented by the fibonacci function.
I have done my base step and elaborated my inductive hypothesis already, but I am having trouble with my inductive step.
$pmb Inductive Hypothesis:$ Let z $in$ $mathbb Z^+$ be given. Assume $sum_i=1^n if_i = nf_n+2 - f_n+3 + 2 $ is true for all n = z.
From what I believe, I must prove that $sum_i=1^z+1 if_i $ = $ (z+1)f_z+3 - f_z+4 + 2$.
I have reached the following:
beginalign
sum_i=1^z+1 if_i &= (sum_i=1^z if_i) + (z+1)f_z+1 \
& = zf_z+2 - f_z+3 + 2 + (z+1)f_z+1 pmb(by inductive hypothesis)
endalign
I do not know where to take it from this point on.
Any help would be greatly appreciated!
discrete-mathematics
discrete-mathematics
asked Mar 29 at 15:07
IOStatIOStat
31
asked Mar 29 at 15:07
IOStatIOStat
31
asked Mar 29 at 15:07
IOStatIOStat
31
asked Mar 29 at 15:07
IOStatIOStat
31
asked Mar 29 at 15:07
IOStatIOStat
31
31
$begingroup$
You need to show it is true for $n=0$, then you should use $f_n+f_n+1=f_n+2$.
$endgroup$
– Alec B-G
Mar 29 at 15:24
add a comment |
$begingroup$
You need to show it is true for $n=0$, then you should use $f_n+f_n+1=f_n+2$.
$endgroup$
– Alec B-G
Mar 29 at 15:24
$begingroup$
You need to show it is true for $n=0$, then you should use $f_n+f_n+1=f_n+2$.
$endgroup$
– Alec B-G
Mar 29 at 15:24
$begingroup$
You need to show it is true for $n=0$, then you should use $f_n+f_n+1=f_n+2$.
$endgroup$
– Alec B-G
Mar 29 at 15:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, for $n=1$, the equality is
$$f_1=f_3-f_4+2$$
This is true because the first terms of the Fibonacci sequence are $f_1 = 1$, $f_2 = 1$, $f_3 = 2$, $f_4 = 3$.
Now, let's suppose that you have, for an integer $n geq 1$, the relation
$$sum_i=1^n if_i = nf_n+2 - f_n+3 + 2$$
Then you have
$$sum_i=1^n+1 if_i = (n+1)f_n+1 + sum_i=1^n if_i = (n+1)f_n+1 + nf_n+2 - f_n+3 + 2$$
(the last step uses induction hypothesis). Using now that $f_n+1 = f_n+3 - f_n+2$, you get
$$sum_i=1^n+1 if_i = nf_n+3 - f_n+2 + 2$$
and using $f_n+2 = f_n+4- f_n+3$, you get
$$sum_i=1^n+1 if_i = (n+1)f_n+3 - f_n+4 + 2$$
i.e. the equality at rank $n+1$. This concludes the induction.
answered Mar 29 at 16:53
TheSilverDoeTheSilverDoe
5,433216
$endgroup$
add a comment |
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1 Answer
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$begingroup$
First, for $n=1$, the equality is
$$f_1=f_3-f_4+2$$
This is true because the first terms of the Fibonacci sequence are $f_1 = 1$, $f_2 = 1$, $f_3 = 2$, $f_4 = 3$.
Now, let's suppose that you have, for an integer $n geq 1$, the relation
$$sum_i=1^n if_i = nf_n+2 - f_n+3 + 2$$
Then you have
$$sum_i=1^n+1 if_i = (n+1)f_n+1 + sum_i=1^n if_i = (n+1)f_n+1 + nf_n+2 - f_n+3 + 2$$
(the last step uses induction hypothesis). Using now that $f_n+1 = f_n+3 - f_n+2$, you get
$$sum_i=1^n+1 if_i = nf_n+3 - f_n+2 + 2$$
and using $f_n+2 = f_n+4- f_n+3$, you get
$$sum_i=1^n+1 if_i = (n+1)f_n+3 - f_n+4 + 2$$
i.e. the equality at rank $n+1$. This concludes the induction.
answered Mar 29 at 16:53
TheSilverDoeTheSilverDoe
5,433216
$endgroup$
add a comment |
$begingroup$
First, for $n=1$, the equality is
$$f_1=f_3-f_4+2$$
This is true because the first terms of the Fibonacci sequence are $f_1 = 1$, $f_2 = 1$, $f_3 = 2$, $f_4 = 3$.
Now, let's suppose that you have, for an integer $n geq 1$, the relation
$$sum_i=1^n if_i = nf_n+2 - f_n+3 + 2$$
Then you have
$$sum_i=1^n+1 if_i = (n+1)f_n+1 + sum_i=1^n if_i = (n+1)f_n+1 + nf_n+2 - f_n+3 + 2$$
(the last step uses induction hypothesis). Using now that $f_n+1 = f_n+3 - f_n+2$, you get
$$sum_i=1^n+1 if_i = nf_n+3 - f_n+2 + 2$$
and using $f_n+2 = f_n+4- f_n+3$, you get
$$sum_i=1^n+1 if_i = (n+1)f_n+3 - f_n+4 + 2$$
i.e. the equality at rank $n+1$. This concludes the induction.
answered Mar 29 at 16:53
TheSilverDoeTheSilverDoe
5,433216
$endgroup$
add a comment |
$begingroup$
First, for $n=1$, the equality is
$$f_1=f_3-f_4+2$$
This is true because the first terms of the Fibonacci sequence are $f_1 = 1$, $f_2 = 1$, $f_3 = 2$, $f_4 = 3$.
Now, let's suppose that you have, for an integer $n geq 1$, the relation
$$sum_i=1^n if_i = nf_n+2 - f_n+3 + 2$$
Then you have
$$sum_i=1^n+1 if_i = (n+1)f_n+1 + sum_i=1^n if_i = (n+1)f_n+1 + nf_n+2 - f_n+3 + 2$$
(the last step uses induction hypothesis). Using now that $f_n+1 = f_n+3 - f_n+2$, you get
$$sum_i=1^n+1 if_i = nf_n+3 - f_n+2 + 2$$
and using $f_n+2 = f_n+4- f_n+3$, you get
$$sum_i=1^n+1 if_i = (n+1)f_n+3 - f_n+4 + 2$$
i.e. the equality at rank $n+1$. This concludes the induction.
answered Mar 29 at 16:53
TheSilverDoeTheSilverDoe
5,433216
$endgroup$
First, for $n=1$, the equality is
$$f_1=f_3-f_4+2$$
This is true because the first terms of the Fibonacci sequence are $f_1 = 1$, $f_2 = 1$, $f_3 = 2$, $f_4 = 3$.
Now, let's suppose that you have, for an integer $n geq 1$, the relation
$$sum_i=1^n if_i = nf_n+2 - f_n+3 + 2$$
Then you have
$$sum_i=1^n+1 if_i = (n+1)f_n+1 + sum_i=1^n if_i = (n+1)f_n+1 + nf_n+2 - f_n+3 + 2$$
(the last step uses induction hypothesis). Using now that $f_n+1 = f_n+3 - f_n+2$, you get
$$sum_i=1^n+1 if_i = nf_n+3 - f_n+2 + 2$$
and using $f_n+2 = f_n+4- f_n+3$, you get
$$sum_i=1^n+1 if_i = (n+1)f_n+3 - f_n+4 + 2$$
i.e. the equality at rank $n+1$. This concludes the induction.
answered Mar 29 at 16:53
TheSilverDoeTheSilverDoe
5,433216
answered Mar 29 at 16:53
TheSilverDoeTheSilverDoe
5,433216
answered Mar 29 at 16:53
TheSilverDoeTheSilverDoe
5,433216
answered Mar 29 at 16:53
TheSilverDoeTheSilverDoe
5,433216
5,433216
add a comment |
add a comment |
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$begingroup$
You need to show it is true for $n=0$, then you should use $f_n+f_n+1=f_n+2$.
$endgroup$
– Alec B-G
Mar 29 at 15:24