Induction Proof for Sequences Announcing the arrival of Valued Associate #679: Cesar Manara ...
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Induction Proof for Sequences
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Find an explicit formula for the recursive sequenceGeneral formula for iterated cumulative sumUpper Bounds ProofShow That a Sequence is Monotonically IncreasingIs there a generalized analogue to the summation and product operators?How to prove this statement with “first” principle of Mathematical induction and not strong Mathematical induction?Sum of infinite series $sum_{i=1}^{infty} frac 1 {i(i+1)(i+2)…(i+n)}$Proof by induction of $s_k=2s_{k-2}$Sum of last digits of a sumProof by Induction: Recursively Defined Sequential Set
$begingroup$
Given a sequence $s_k=s_{k-1}+6k$, where $s_0=7$.
Question: First, find the closed formula for the $n$-th component of this sequence by hand and then prove that your formula is correct
My attempt:
I found the first couple of terms of the sequence to be
$s_0=7$,
$s_1=13$,
$s_2=25$,
$s_3=43$,
$s_4=67$ and
$s_5=97$.
I found the formula for the $n$-th term to be $s_n=3n^2+3n+7$.
Proof:
Base case $s_0=7$ therefore
$7=3cdot(0)^2+3cdot(0)+7$ so the formula works for the $s_0$ element.
I'm not sure how to proceed from here but I believe the proof should be a proof by Strong Induction. Any help will be greatly appreciated.
sequences-and-series induction
New contributor
$endgroup$
add a comment |
$begingroup$
Given a sequence $s_k=s_{k-1}+6k$, where $s_0=7$.
Question: First, find the closed formula for the $n$-th component of this sequence by hand and then prove that your formula is correct
My attempt:
I found the first couple of terms of the sequence to be
$s_0=7$,
$s_1=13$,
$s_2=25$,
$s_3=43$,
$s_4=67$ and
$s_5=97$.
I found the formula for the $n$-th term to be $s_n=3n^2+3n+7$.
Proof:
Base case $s_0=7$ therefore
$7=3cdot(0)^2+3cdot(0)+7$ so the formula works for the $s_0$ element.
I'm not sure how to proceed from here but I believe the proof should be a proof by Strong Induction. Any help will be greatly appreciated.
sequences-and-series induction
New contributor
$endgroup$
add a comment |
$begingroup$
Given a sequence $s_k=s_{k-1}+6k$, where $s_0=7$.
Question: First, find the closed formula for the $n$-th component of this sequence by hand and then prove that your formula is correct
My attempt:
I found the first couple of terms of the sequence to be
$s_0=7$,
$s_1=13$,
$s_2=25$,
$s_3=43$,
$s_4=67$ and
$s_5=97$.
I found the formula for the $n$-th term to be $s_n=3n^2+3n+7$.
Proof:
Base case $s_0=7$ therefore
$7=3cdot(0)^2+3cdot(0)+7$ so the formula works for the $s_0$ element.
I'm not sure how to proceed from here but I believe the proof should be a proof by Strong Induction. Any help will be greatly appreciated.
sequences-and-series induction
New contributor
$endgroup$
Given a sequence $s_k=s_{k-1}+6k$, where $s_0=7$.
Question: First, find the closed formula for the $n$-th component of this sequence by hand and then prove that your formula is correct
My attempt:
I found the first couple of terms of the sequence to be
$s_0=7$,
$s_1=13$,
$s_2=25$,
$s_3=43$,
$s_4=67$ and
$s_5=97$.
I found the formula for the $n$-th term to be $s_n=3n^2+3n+7$.
Proof:
Base case $s_0=7$ therefore
$7=3cdot(0)^2+3cdot(0)+7$ so the formula works for the $s_0$ element.
I'm not sure how to proceed from here but I believe the proof should be a proof by Strong Induction. Any help will be greatly appreciated.
sequences-and-series induction
sequences-and-series induction
New contributor
New contributor
edited 6 hours ago
Marian G.
38426
38426
New contributor
asked 6 hours ago
JuliaJulia
113
113
New contributor
New contributor
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You just apply the recurrence
to your induction hypothesis.
If
$s_n=3n^2+3n+7
$,
then,
since
$s_k=s_{k-1}+6k
$,
$begin{array}\
s_{n+1}
&=s_n+6(n+1)\
&=3n^2+3n+7+6(n+1)\
&=3n^2+9n+13\
end{array}
$
If your hypothesis is true,
then
$begin{array}\
s_{n+1}
&=3(n+1)^2+3(n+1)+7\
&=3(n^2+2n+1)+3(n+1)+7\
&=3n^2+6n+3+3n+3+7\
&=3n^2+9n+13\
end{array}
$
This matches the result
from the induction step,
so the induction hypothesis is proved.
$endgroup$
add a comment |
$begingroup$
FYI, here is a way to determine what the closed formula is without having to determine it by hand. This is an extension of the answer given by szw1710. The given sequence is
$$s_k=s_{k-1}+6k tag{1}label{eq1}$$
Summing both sides of eqref{eq1} from $1$ to $n$ gives that
$sum_{k=1}^{n}s_k = sum_{k=1}^{n}s_{k-1} + sum_{k=1}^{n}6k$
$sum_{k=1}^{n-1}s_k + s_n = s_0 + sum_{k=1}^{n-1}s_{k} + 6frac{n(n+1)}{2}$
$s_n = 7 + 3n(n+1) = 3n^2 + 3n + 7$
As you can see, this technique can easily be used in any cases where you have $s_k = s_{k-1} + f(k)$ and the sum of $f(k)$ up to $k = n$ can be fairly easily determined.
More generally, your question is a fairly simple example of Linear Recurrence Relations with Constant Coefficients. You can use a certain technique of a characteristic equation, as described in that link, to directly determine the solution of even considerably more complicated such equations.
$endgroup$
add a comment |
$begingroup$
It seems to me that induction is not needed here. Fix $kinBbb N.$ A direct computation shows that $s_k-s_{k-1}=6k$ and $s_0=7.$ However, there is another problem: both induction, and my method show that if $s_k=3k^2+3k+7$, then $s_{k}=s_{k-1}+6k$. In fact, whe should prove the converse: if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $s_k=3k^2+3k+7$. I will look for some reasoning going in this direction.
Let $s_0=7$ and $s_{k}=s_{k-1}+6k.$ Define $t_k=s_k-3k^2-3k-7.$ Then $t_0=0$. It is easy to prove (by direct computation) that $t_k=t_{k-1}$, so $(t_k)$ is a constant (in fact, zero) sequence. Then $s_k=3k^2+3k+7.$
$endgroup$
1
$begingroup$
My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
$endgroup$
– John Omielan
4 hours ago
$begingroup$
Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
$endgroup$
– szw1710
4 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You just apply the recurrence
to your induction hypothesis.
If
$s_n=3n^2+3n+7
$,
then,
since
$s_k=s_{k-1}+6k
$,
$begin{array}\
s_{n+1}
&=s_n+6(n+1)\
&=3n^2+3n+7+6(n+1)\
&=3n^2+9n+13\
end{array}
$
If your hypothesis is true,
then
$begin{array}\
s_{n+1}
&=3(n+1)^2+3(n+1)+7\
&=3(n^2+2n+1)+3(n+1)+7\
&=3n^2+6n+3+3n+3+7\
&=3n^2+9n+13\
end{array}
$
This matches the result
from the induction step,
so the induction hypothesis is proved.
$endgroup$
add a comment |
$begingroup$
You just apply the recurrence
to your induction hypothesis.
If
$s_n=3n^2+3n+7
$,
then,
since
$s_k=s_{k-1}+6k
$,
$begin{array}\
s_{n+1}
&=s_n+6(n+1)\
&=3n^2+3n+7+6(n+1)\
&=3n^2+9n+13\
end{array}
$
If your hypothesis is true,
then
$begin{array}\
s_{n+1}
&=3(n+1)^2+3(n+1)+7\
&=3(n^2+2n+1)+3(n+1)+7\
&=3n^2+6n+3+3n+3+7\
&=3n^2+9n+13\
end{array}
$
This matches the result
from the induction step,
so the induction hypothesis is proved.
$endgroup$
add a comment |
$begingroup$
You just apply the recurrence
to your induction hypothesis.
If
$s_n=3n^2+3n+7
$,
then,
since
$s_k=s_{k-1}+6k
$,
$begin{array}\
s_{n+1}
&=s_n+6(n+1)\
&=3n^2+3n+7+6(n+1)\
&=3n^2+9n+13\
end{array}
$
If your hypothesis is true,
then
$begin{array}\
s_{n+1}
&=3(n+1)^2+3(n+1)+7\
&=3(n^2+2n+1)+3(n+1)+7\
&=3n^2+6n+3+3n+3+7\
&=3n^2+9n+13\
end{array}
$
This matches the result
from the induction step,
so the induction hypothesis is proved.
$endgroup$
You just apply the recurrence
to your induction hypothesis.
If
$s_n=3n^2+3n+7
$,
then,
since
$s_k=s_{k-1}+6k
$,
$begin{array}\
s_{n+1}
&=s_n+6(n+1)\
&=3n^2+3n+7+6(n+1)\
&=3n^2+9n+13\
end{array}
$
If your hypothesis is true,
then
$begin{array}\
s_{n+1}
&=3(n+1)^2+3(n+1)+7\
&=3(n^2+2n+1)+3(n+1)+7\
&=3n^2+6n+3+3n+3+7\
&=3n^2+9n+13\
end{array}
$
This matches the result
from the induction step,
so the induction hypothesis is proved.
answered 6 hours ago
marty cohenmarty cohen
75.9k549130
75.9k549130
add a comment |
add a comment |
$begingroup$
FYI, here is a way to determine what the closed formula is without having to determine it by hand. This is an extension of the answer given by szw1710. The given sequence is
$$s_k=s_{k-1}+6k tag{1}label{eq1}$$
Summing both sides of eqref{eq1} from $1$ to $n$ gives that
$sum_{k=1}^{n}s_k = sum_{k=1}^{n}s_{k-1} + sum_{k=1}^{n}6k$
$sum_{k=1}^{n-1}s_k + s_n = s_0 + sum_{k=1}^{n-1}s_{k} + 6frac{n(n+1)}{2}$
$s_n = 7 + 3n(n+1) = 3n^2 + 3n + 7$
As you can see, this technique can easily be used in any cases where you have $s_k = s_{k-1} + f(k)$ and the sum of $f(k)$ up to $k = n$ can be fairly easily determined.
More generally, your question is a fairly simple example of Linear Recurrence Relations with Constant Coefficients. You can use a certain technique of a characteristic equation, as described in that link, to directly determine the solution of even considerably more complicated such equations.
$endgroup$
add a comment |
$begingroup$
FYI, here is a way to determine what the closed formula is without having to determine it by hand. This is an extension of the answer given by szw1710. The given sequence is
$$s_k=s_{k-1}+6k tag{1}label{eq1}$$
Summing both sides of eqref{eq1} from $1$ to $n$ gives that
$sum_{k=1}^{n}s_k = sum_{k=1}^{n}s_{k-1} + sum_{k=1}^{n}6k$
$sum_{k=1}^{n-1}s_k + s_n = s_0 + sum_{k=1}^{n-1}s_{k} + 6frac{n(n+1)}{2}$
$s_n = 7 + 3n(n+1) = 3n^2 + 3n + 7$
As you can see, this technique can easily be used in any cases where you have $s_k = s_{k-1} + f(k)$ and the sum of $f(k)$ up to $k = n$ can be fairly easily determined.
More generally, your question is a fairly simple example of Linear Recurrence Relations with Constant Coefficients. You can use a certain technique of a characteristic equation, as described in that link, to directly determine the solution of even considerably more complicated such equations.
$endgroup$
add a comment |
$begingroup$
FYI, here is a way to determine what the closed formula is without having to determine it by hand. This is an extension of the answer given by szw1710. The given sequence is
$$s_k=s_{k-1}+6k tag{1}label{eq1}$$
Summing both sides of eqref{eq1} from $1$ to $n$ gives that
$sum_{k=1}^{n}s_k = sum_{k=1}^{n}s_{k-1} + sum_{k=1}^{n}6k$
$sum_{k=1}^{n-1}s_k + s_n = s_0 + sum_{k=1}^{n-1}s_{k} + 6frac{n(n+1)}{2}$
$s_n = 7 + 3n(n+1) = 3n^2 + 3n + 7$
As you can see, this technique can easily be used in any cases where you have $s_k = s_{k-1} + f(k)$ and the sum of $f(k)$ up to $k = n$ can be fairly easily determined.
More generally, your question is a fairly simple example of Linear Recurrence Relations with Constant Coefficients. You can use a certain technique of a characteristic equation, as described in that link, to directly determine the solution of even considerably more complicated such equations.
$endgroup$
FYI, here is a way to determine what the closed formula is without having to determine it by hand. This is an extension of the answer given by szw1710. The given sequence is
$$s_k=s_{k-1}+6k tag{1}label{eq1}$$
Summing both sides of eqref{eq1} from $1$ to $n$ gives that
$sum_{k=1}^{n}s_k = sum_{k=1}^{n}s_{k-1} + sum_{k=1}^{n}6k$
$sum_{k=1}^{n-1}s_k + s_n = s_0 + sum_{k=1}^{n-1}s_{k} + 6frac{n(n+1)}{2}$
$s_n = 7 + 3n(n+1) = 3n^2 + 3n + 7$
As you can see, this technique can easily be used in any cases where you have $s_k = s_{k-1} + f(k)$ and the sum of $f(k)$ up to $k = n$ can be fairly easily determined.
More generally, your question is a fairly simple example of Linear Recurrence Relations with Constant Coefficients. You can use a certain technique of a characteristic equation, as described in that link, to directly determine the solution of even considerably more complicated such equations.
edited 4 hours ago
answered 4 hours ago
John OmielanJohn Omielan
5,2192218
5,2192218
add a comment |
add a comment |
$begingroup$
It seems to me that induction is not needed here. Fix $kinBbb N.$ A direct computation shows that $s_k-s_{k-1}=6k$ and $s_0=7.$ However, there is another problem: both induction, and my method show that if $s_k=3k^2+3k+7$, then $s_{k}=s_{k-1}+6k$. In fact, whe should prove the converse: if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $s_k=3k^2+3k+7$. I will look for some reasoning going in this direction.
Let $s_0=7$ and $s_{k}=s_{k-1}+6k.$ Define $t_k=s_k-3k^2-3k-7.$ Then $t_0=0$. It is easy to prove (by direct computation) that $t_k=t_{k-1}$, so $(t_k)$ is a constant (in fact, zero) sequence. Then $s_k=3k^2+3k+7.$
$endgroup$
1
$begingroup$
My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
$endgroup$
– John Omielan
4 hours ago
$begingroup$
Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
$endgroup$
– szw1710
4 hours ago
add a comment |
$begingroup$
It seems to me that induction is not needed here. Fix $kinBbb N.$ A direct computation shows that $s_k-s_{k-1}=6k$ and $s_0=7.$ However, there is another problem: both induction, and my method show that if $s_k=3k^2+3k+7$, then $s_{k}=s_{k-1}+6k$. In fact, whe should prove the converse: if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $s_k=3k^2+3k+7$. I will look for some reasoning going in this direction.
Let $s_0=7$ and $s_{k}=s_{k-1}+6k.$ Define $t_k=s_k-3k^2-3k-7.$ Then $t_0=0$. It is easy to prove (by direct computation) that $t_k=t_{k-1}$, so $(t_k)$ is a constant (in fact, zero) sequence. Then $s_k=3k^2+3k+7.$
$endgroup$
1
$begingroup$
My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
$endgroup$
– John Omielan
4 hours ago
$begingroup$
Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
$endgroup$
– szw1710
4 hours ago
add a comment |
$begingroup$
It seems to me that induction is not needed here. Fix $kinBbb N.$ A direct computation shows that $s_k-s_{k-1}=6k$ and $s_0=7.$ However, there is another problem: both induction, and my method show that if $s_k=3k^2+3k+7$, then $s_{k}=s_{k-1}+6k$. In fact, whe should prove the converse: if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $s_k=3k^2+3k+7$. I will look for some reasoning going in this direction.
Let $s_0=7$ and $s_{k}=s_{k-1}+6k.$ Define $t_k=s_k-3k^2-3k-7.$ Then $t_0=0$. It is easy to prove (by direct computation) that $t_k=t_{k-1}$, so $(t_k)$ is a constant (in fact, zero) sequence. Then $s_k=3k^2+3k+7.$
$endgroup$
It seems to me that induction is not needed here. Fix $kinBbb N.$ A direct computation shows that $s_k-s_{k-1}=6k$ and $s_0=7.$ However, there is another problem: both induction, and my method show that if $s_k=3k^2+3k+7$, then $s_{k}=s_{k-1}+6k$. In fact, whe should prove the converse: if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $s_k=3k^2+3k+7$. I will look for some reasoning going in this direction.
Let $s_0=7$ and $s_{k}=s_{k-1}+6k.$ Define $t_k=s_k-3k^2-3k-7.$ Then $t_0=0$. It is easy to prove (by direct computation) that $t_k=t_{k-1}$, so $(t_k)$ is a constant (in fact, zero) sequence. Then $s_k=3k^2+3k+7.$
edited 4 hours ago
answered 6 hours ago
szw1710szw1710
6,5701223
6,5701223
1
$begingroup$
My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
$endgroup$
– John Omielan
4 hours ago
$begingroup$
Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
$endgroup$
– szw1710
4 hours ago
add a comment |
1
$begingroup$
My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
$endgroup$
– John Omielan
4 hours ago
$begingroup$
Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
$endgroup$
– szw1710
4 hours ago
1
1
$begingroup$
My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
$endgroup$
– John Omielan
4 hours ago
$begingroup$
My answer has one way to show what you're asking about, i.e., "if $s_{k}=s_{k-1}+6k$ and $s_0=7$, then $a_k=3k^2+3k+7$".
$endgroup$
– John Omielan
4 hours ago
$begingroup$
Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
$endgroup$
– szw1710
4 hours ago
$begingroup$
Yes, of course. About such techniques one could read in the "Concrete mathematics" book by Graham, Knuth, Patashnik.
$endgroup$
– szw1710
4 hours ago
add a comment |
Julia is a new contributor. Be nice, and check out our Code of Conduct.
Julia is a new contributor. Be nice, and check out our Code of Conduct.
Julia is a new contributor. Be nice, and check out our Code of Conduct.
Julia is a new contributor. Be nice, and check out our Code of Conduct.
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