Mixing
An Induction Problem, What Am I Supposed To Prove?
Clash Royale CLAN TAG#URR8PPP
up vote
7
down vote
favorite
I have encountered an induction problem which I don't understand. What I don't understand is what it is asking me to prove. I don't want a solution. The problem is:
If $u_1=5$ and $u_n+1=2u_n-3(-1)^n$, then $u_n=3(2^n)+(-1)^n$ for all positive integers.
Am I supposed to prove $u_n+1=2u_n-3(-1)^n$ or $u_n=3(2^n)+(-1)^n$ is true for all positive integers?
induction
edited Sep 7 at 14:28
Babelfish
1,004115
asked Sep 7 at 14:11
sup
864
New contributor
sup is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
7
down vote
favorite
I have encountered an induction problem which I don't understand. What I don't understand is what it is asking me to prove. I don't want a solution. The problem is:
If $u_1=5$ and $u_n+1=2u_n-3(-1)^n$, then $u_n=3(2^n)+(-1)^n$ for all positive integers.
Am I supposed to prove $u_n+1=2u_n-3(-1)^n$ or $u_n=3(2^n)+(-1)^n$ is true for all positive integers?
induction
edited Sep 7 at 14:28
Babelfish
1,004115
asked Sep 7 at 14:11
sup
864
New contributor
sup is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
You are supposed to prove that $u_n= 3(2^n)+(-1)^n$. $u_n+1=2u_n-3(-1)^n$ is a definition.
– m_t_
Sep 7 at 14:15
Induction is : [if $P(1)$ and "if $P(n)$, then $P(n+1)$, for every $n$", then "for every $n$, $P(n)$"].
– Mauro ALLEGRANZA
Sep 7 at 14:16
The expression for $u_n$ is the $P(n)$. First step: is it true that for $n=1$ its value is $5$ ?
– Mauro ALLEGRANZA
Sep 7 at 14:17
4
$",colorbluestylefont-family:inherittextIf;u_1=5 ;stylefont-family:inherittextand ;u_n+1=2u_n-3(-1)^n,, ;colorredstylefont-family:inherittext then;u_n=3(2^n)+(-1)^n,",$ means you are given the blue part, and have to prove the red part.
– dxiv
Sep 7 at 19:41
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I have encountered an induction problem which I don't understand. What I don't understand is what it is asking me to prove. I don't want a solution. The problem is:
If $u_1=5$ and $u_n+1=2u_n-3(-1)^n$, then $u_n=3(2^n)+(-1)^n$ for all positive integers.
Am I supposed to prove $u_n+1=2u_n-3(-1)^n$ or $u_n=3(2^n)+(-1)^n$ is true for all positive integers?
induction
edited Sep 7 at 14:28
Babelfish
1,004115
asked Sep 7 at 14:11
sup
864
New contributor
sup is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have encountered an induction problem which I don't understand. What I don't understand is what it is asking me to prove. I don't want a solution. The problem is:
If $u_1=5$ and $u_n+1=2u_n-3(-1)^n$, then $u_n=3(2^n)+(-1)^n$ for all positive integers.
Am I supposed to prove $u_n+1=2u_n-3(-1)^n$ or $u_n=3(2^n)+(-1)^n$ is true for all positive integers?
induction
edited Sep 7 at 14:28
Babelfish
1,004115
asked Sep 7 at 14:11
sup
864
New contributor
sup is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Sep 7 at 14:28
Babelfish
1,004115
edited Sep 7 at 14:28
Babelfish
1,004115
edited Sep 7 at 14:28
Babelfish
1,004115
1,004115
asked Sep 7 at 14:11
sup
864
New contributor
sup is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Sep 7 at 14:11
sup
864
asked Sep 7 at 14:11
sup
864
864
New contributor
sup is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
sup is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
sup is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
You are supposed to prove that $u_n= 3(2^n)+(-1)^n$. $u_n+1=2u_n-3(-1)^n$ is a definition.
– m_t_
Sep 7 at 14:15
Induction is : [if $P(1)$ and "if $P(n)$, then $P(n+1)$, for every $n$", then "for every $n$, $P(n)$"].
– Mauro ALLEGRANZA
Sep 7 at 14:16
The expression for $u_n$ is the $P(n)$. First step: is it true that for $n=1$ its value is $5$ ?
– Mauro ALLEGRANZA
Sep 7 at 14:17
4
$",colorbluestylefont-family:inherittextIf;u_1=5 ;stylefont-family:inherittextand ;u_n+1=2u_n-3(-1)^n,, ;colorredstylefont-family:inherittext then;u_n=3(2^n)+(-1)^n,",$ means you are given the blue part, and have to prove the red part.
– dxiv
Sep 7 at 19:41
add a comment |Â
3
You are supposed to prove that $u_n= 3(2^n)+(-1)^n$. $u_n+1=2u_n-3(-1)^n$ is a definition.
– m_t_
Sep 7 at 14:15
Induction is : [if $P(1)$ and "if $P(n)$, then $P(n+1)$, for every $n$", then "for every $n$, $P(n)$"].
– Mauro ALLEGRANZA
Sep 7 at 14:16
The expression for $u_n$ is the $P(n)$. First step: is it true that for $n=1$ its value is $5$ ?
– Mauro ALLEGRANZA
Sep 7 at 14:17
4
$",colorbluestylefont-family:inherittextIf;u_1=5 ;stylefont-family:inherittextand ;u_n+1=2u_n-3(-1)^n,, ;colorredstylefont-family:inherittext then;u_n=3(2^n)+(-1)^n,",$ means you are given the blue part, and have to prove the red part.
– dxiv
Sep 7 at 19:41
3
3
You are supposed to prove that $u_n= 3(2^n)+(-1)^n$. $u_n+1=2u_n-3(-1)^n$ is a definition.
– m_t_
Sep 7 at 14:15
You are supposed to prove that $u_n= 3(2^n)+(-1)^n$. $u_n+1=2u_n-3(-1)^n$ is a definition.
– m_t_
Sep 7 at 14:15
Induction is : [if $P(1)$ and "if $P(n)$, then $P(n+1)$, for every $n$", then "for every $n$, $P(n)$"].
– Mauro ALLEGRANZA
Sep 7 at 14:16
Induction is : [if $P(1)$ and "if $P(n)$, then $P(n+1)$, for every $n$", then "for every $n$, $P(n)$"].
– Mauro ALLEGRANZA
Sep 7 at 14:16
The expression for $u_n$ is the $P(n)$. First step: is it true that for $n=1$ its value is $5$ ?
– Mauro ALLEGRANZA
Sep 7 at 14:17
The expression for $u_n$ is the $P(n)$. First step: is it true that for $n=1$ its value is $5$ ?
– Mauro ALLEGRANZA
Sep 7 at 14:17
4
4
$",colorbluestylefont-family:inherittextIf;u_1=5 ;stylefont-family:inherittextand ;u_n+1=2u_n-3(-1)^n,, ;colorredstylefont-family:inherittext then;u_n=3(2^n)+(-1)^n,",$ means you are given the blue part, and have to prove the red part.
– dxiv
Sep 7 at 19:41
$",colorbluestylefont-family:inherittextIf;u_1=5 ;stylefont-family:inherittextand ;u_n+1=2u_n-3(-1)^n,, ;colorredstylefont-family:inherittext then;u_n=3(2^n)+(-1)^n,",$ means you are given the blue part, and have to prove the red part.
– dxiv
Sep 7 at 19:41
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
9
down vote
accepted
You are supposed to prove $u_n=3(2^n)+(-1)^n$.
$u_1=5$ and $u_n+1=2u_n-3(-1)^n$ are the conditions you are supposed to make use of.
answered Sep 7 at 14:15
Yuta
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
$u_n+1=2u_n-3(-1)^n$ is your recurrence, specifying on how to obtain the $n+1$-st term $u_n+1$ from the $n$-th term $u_n$. Together with a start value, $u_1=5$, this stepwise determines the sequence completely.
What the problem is trying to establish, is to show that in general, for any $n$, you can obtain the $n$-th term directly(without evaluating all $u_k$'s with $k<n$ before) via the formula $u_n=3(2^n)+(-1)^n$. That this formula holds can be shown using induction.
answered Sep 7 at 14:16
zzuussee
2,480625
add a comment |Â
up vote
0
down vote
You are given $u_n+1=2u_n-3(-1)^n$. This would allow you to compute the entire series. For example $$u_2=2u_1-3(-1)^1=2cdot 5+3=13\ u_3=2cdot 13-3(-1)^2=23$$ and so on. You are supposed to prove the last statement. It claims
$$u_1=3(2^1)+(-1)^1=5\u_2=3(2^2)+(-1)^2=13\u_3=3(2^3)+(-1)^3=23\vdots$$
edited Sep 8 at 6:26
user21820
36.2k440140
answered Sep 7 at 14:23
Ross Millikan
279k23189355
What do you mean it's incorrect?
– sup
Sep 7 at 14:25
I found my problem. I had the wrong exponent on the $(-1)^n$
– Ross Millikan
Sep 7 at 14:27
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
You are supposed to prove $u_n=3(2^n)+(-1)^n$.
$u_1=5$ and $u_n+1=2u_n-3(-1)^n$ are the conditions you are supposed to make use of.
answered Sep 7 at 14:15
Yuta
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
9
down vote
accepted
You are supposed to prove $u_n=3(2^n)+(-1)^n$.
$u_1=5$ and $u_n+1=2u_n-3(-1)^n$ are the conditions you are supposed to make use of.
answered Sep 7 at 14:15
Yuta
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
You are supposed to prove $u_n=3(2^n)+(-1)^n$.
$u_1=5$ and $u_n+1=2u_n-3(-1)^n$ are the conditions you are supposed to make use of.
answered Sep 7 at 14:15
Yuta
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You are supposed to prove $u_n=3(2^n)+(-1)^n$.
$u_1=5$ and $u_n+1=2u_n-3(-1)^n$ are the conditions you are supposed to make use of.
answered Sep 7 at 14:15
Yuta
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Sep 7 at 14:15
Yuta
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Sep 7 at 14:15
Yuta
62929
answered Sep 7 at 14:15
Yuta
62929
62929
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
2
down vote
$u_n+1=2u_n-3(-1)^n$ is your recurrence, specifying on how to obtain the $n+1$-st term $u_n+1$ from the $n$-th term $u_n$. Together with a start value, $u_1=5$, this stepwise determines the sequence completely.
What the problem is trying to establish, is to show that in general, for any $n$, you can obtain the $n$-th term directly(without evaluating all $u_k$'s with $k<n$ before) via the formula $u_n=3(2^n)+(-1)^n$. That this formula holds can be shown using induction.
answered Sep 7 at 14:16
zzuussee
2,480625
add a comment |Â
up vote
2
down vote
$u_n+1=2u_n-3(-1)^n$ is your recurrence, specifying on how to obtain the $n+1$-st term $u_n+1$ from the $n$-th term $u_n$. Together with a start value, $u_1=5$, this stepwise determines the sequence completely.
What the problem is trying to establish, is to show that in general, for any $n$, you can obtain the $n$-th term directly(without evaluating all $u_k$'s with $k<n$ before) via the formula $u_n=3(2^n)+(-1)^n$. That this formula holds can be shown using induction.
answered Sep 7 at 14:16
zzuussee
2,480625
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$u_n+1=2u_n-3(-1)^n$ is your recurrence, specifying on how to obtain the $n+1$-st term $u_n+1$ from the $n$-th term $u_n$. Together with a start value, $u_1=5$, this stepwise determines the sequence completely.
What the problem is trying to establish, is to show that in general, for any $n$, you can obtain the $n$-th term directly(without evaluating all $u_k$'s with $k<n$ before) via the formula $u_n=3(2^n)+(-1)^n$. That this formula holds can be shown using induction.
answered Sep 7 at 14:16
zzuussee
2,480625
$u_n+1=2u_n-3(-1)^n$ is your recurrence, specifying on how to obtain the $n+1$-st term $u_n+1$ from the $n$-th term $u_n$. Together with a start value, $u_1=5$, this stepwise determines the sequence completely.
What the problem is trying to establish, is to show that in general, for any $n$, you can obtain the $n$-th term directly(without evaluating all $u_k$'s with $k<n$ before) via the formula $u_n=3(2^n)+(-1)^n$. That this formula holds can be shown using induction.
answered Sep 7 at 14:16
zzuussee
2,480625
answered Sep 7 at 14:16
zzuussee
2,480625
answered Sep 7 at 14:16
zzuussee
2,480625
answered Sep 7 at 14:16
zzuussee
2,480625
2,480625
add a comment |Â
add a comment |Â
up vote
0
down vote
You are given $u_n+1=2u_n-3(-1)^n$. This would allow you to compute the entire series. For example $$u_2=2u_1-3(-1)^1=2cdot 5+3=13\ u_3=2cdot 13-3(-1)^2=23$$ and so on. You are supposed to prove the last statement. It claims
$$u_1=3(2^1)+(-1)^1=5\u_2=3(2^2)+(-1)^2=13\u_3=3(2^3)+(-1)^3=23\vdots$$
edited Sep 8 at 6:26
user21820
36.2k440140
answered Sep 7 at 14:23
Ross Millikan
279k23189355
What do you mean it's incorrect?
– sup
Sep 7 at 14:25
I found my problem. I had the wrong exponent on the $(-1)^n$
– Ross Millikan
Sep 7 at 14:27
add a comment |Â
up vote
0
down vote
You are given $u_n+1=2u_n-3(-1)^n$. This would allow you to compute the entire series. For example $$u_2=2u_1-3(-1)^1=2cdot 5+3=13\ u_3=2cdot 13-3(-1)^2=23$$ and so on. You are supposed to prove the last statement. It claims
$$u_1=3(2^1)+(-1)^1=5\u_2=3(2^2)+(-1)^2=13\u_3=3(2^3)+(-1)^3=23\vdots$$
edited Sep 8 at 6:26
user21820
36.2k440140
answered Sep 7 at 14:23
Ross Millikan
279k23189355
What do you mean it's incorrect?
– sup
Sep 7 at 14:25
I found my problem. I had the wrong exponent on the $(-1)^n$
– Ross Millikan
Sep 7 at 14:27
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You are given $u_n+1=2u_n-3(-1)^n$. This would allow you to compute the entire series. For example $$u_2=2u_1-3(-1)^1=2cdot 5+3=13\ u_3=2cdot 13-3(-1)^2=23$$ and so on. You are supposed to prove the last statement. It claims
$$u_1=3(2^1)+(-1)^1=5\u_2=3(2^2)+(-1)^2=13\u_3=3(2^3)+(-1)^3=23\vdots$$
edited Sep 8 at 6:26
user21820
36.2k440140
answered Sep 7 at 14:23
Ross Millikan
279k23189355
You are given $u_n+1=2u_n-3(-1)^n$. This would allow you to compute the entire series. For example $$u_2=2u_1-3(-1)^1=2cdot 5+3=13\ u_3=2cdot 13-3(-1)^2=23$$ and so on. You are supposed to prove the last statement. It claims
$$u_1=3(2^1)+(-1)^1=5\u_2=3(2^2)+(-1)^2=13\u_3=3(2^3)+(-1)^3=23\vdots$$
edited Sep 8 at 6:26
user21820
36.2k440140
answered Sep 7 at 14:23
Ross Millikan
279k23189355
edited Sep 8 at 6:26
user21820
36.2k440140
edited Sep 8 at 6:26
user21820
36.2k440140
edited Sep 8 at 6:26
user21820
36.2k440140
36.2k440140
answered Sep 7 at 14:23
Ross Millikan
279k23189355
answered Sep 7 at 14:23
Ross Millikan
279k23189355
answered Sep 7 at 14:23
Ross Millikan
279k23189355
279k23189355
What do you mean it's incorrect?
– sup
Sep 7 at 14:25
I found my problem. I had the wrong exponent on the $(-1)^n$
– Ross Millikan
Sep 7 at 14:27
add a comment |Â
What do you mean it's incorrect?
– sup
Sep 7 at 14:25
I found my problem. I had the wrong exponent on the $(-1)^n$
– Ross Millikan
Sep 7 at 14:27
What do you mean it's incorrect?
– sup
Sep 7 at 14:25
What do you mean it's incorrect?
– sup
Sep 7 at 14:25
I found my problem. I had the wrong exponent on the $(-1)^n$
– Ross Millikan
Sep 7 at 14:27
I found my problem. I had the wrong exponent on the $(-1)^n$
– Ross Millikan
Sep 7 at 14:27
add a comment |Â
sup is a new contributor. Be nice, and check out our Code of Conduct.
sup is a new contributor. Be nice, and check out our Code of Conduct.
sup is a new contributor. Be nice, and check out our Code of Conduct.
sup is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2908655%2fan-induction-problem-what-am-i-supposed-to-prove%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
3
You are supposed to prove that $u_n= 3(2^n)+(-1)^n$. $u_n+1=2u_n-3(-1)^n$ is a definition.
– m_t_
Sep 7 at 14:15
Induction is : [if $P(1)$ and "if $P(n)$, then $P(n+1)$, for every $n$", then "for every $n$, $P(n)$"].
– Mauro ALLEGRANZA
Sep 7 at 14:16
The expression for $u_n$ is the $P(n)$. First step: is it true that for $n=1$ its value is $5$ ?
– Mauro ALLEGRANZA
Sep 7 at 14:17
4
$",colorbluestylefont-family:inherittextIf;u_1=5 ;stylefont-family:inherittextand ;u_n+1=2u_n-3(-1)^n,, ;colorredstylefont-family:inherittext then;u_n=3(2^n)+(-1)^n,",$ means you are given the blue part, and have to prove the red part.
– dxiv
Sep 7 at 19:41