Solving a series with this form without using induction or countdown method
Ok so i have to solve the question above and find the general term of this sequence without using the induction method or the countdown method. So i want to calcul this with the geometric formula or arithmethic given :
I thought this would be a summation of an arithmetic sequence but using the formula i don't get the answer.
I've tested the answer with wolframalpha entering : y(0)=2,y(n)=2y(n-1)+3n-4
Using the geometric formula i get close to the answer but missing -3n, so i'm kinda lost. Could anyone help thanks !
Maybe it can't be solved with geometric or arithmetic formula and i'd have to use an other method ?
sequences-and-series discrete-mathematics summation
edited Nov 29 '18 at 8:58
dkaeae
304311
asked Nov 28 '18 at 15:30
Dany Pépin
112
add a comment |
Ok so i have to solve the question above and find the general term of this sequence without using the induction method or the countdown method. So i want to calcul this with the geometric formula or arithmethic given :
I thought this would be a summation of an arithmetic sequence but using the formula i don't get the answer.
I've tested the answer with wolframalpha entering : y(0)=2,y(n)=2y(n-1)+3n-4
Using the geometric formula i get close to the answer but missing -3n, so i'm kinda lost. Could anyone help thanks !
Maybe it can't be solved with geometric or arithmetic formula and i'd have to use an other method ?
sequences-and-series discrete-mathematics summation
edited Nov 29 '18 at 8:58
dkaeae
304311
asked Nov 28 '18 at 15:30
Dany Pépin
112
add a comment |
Ok so i have to solve the question above and find the general term of this sequence without using the induction method or the countdown method. So i want to calcul this with the geometric formula or arithmethic given :
I thought this would be a summation of an arithmetic sequence but using the formula i don't get the answer.
I've tested the answer with wolframalpha entering : y(0)=2,y(n)=2y(n-1)+3n-4
Using the geometric formula i get close to the answer but missing -3n, so i'm kinda lost. Could anyone help thanks !
Maybe it can't be solved with geometric or arithmetic formula and i'd have to use an other method ?
sequences-and-series discrete-mathematics summation
edited Nov 29 '18 at 8:58
dkaeae
304311
asked Nov 28 '18 at 15:30
Dany Pépin
112
Ok so i have to solve the question above and find the general term of this sequence without using the induction method or the countdown method. So i want to calcul this with the geometric formula or arithmethic given :
I thought this would be a summation of an arithmetic sequence but using the formula i don't get the answer.
I've tested the answer with wolframalpha entering : y(0)=2,y(n)=2y(n-1)+3n-4
Using the geometric formula i get close to the answer but missing -3n, so i'm kinda lost. Could anyone help thanks !
Maybe it can't be solved with geometric or arithmetic formula and i'd have to use an other method ?
sequences-and-series discrete-mathematics summation
sequences-and-series discrete-mathematics summation
edited Nov 29 '18 at 8:58
dkaeae
304311
asked Nov 28 '18 at 15:30
Dany Pépin
112
edited Nov 29 '18 at 8:58
dkaeae
304311
asked Nov 28 '18 at 15:30
Dany Pépin
112
edited Nov 29 '18 at 8:58
dkaeae
304311
edited Nov 29 '18 at 8:58
dkaeae
304311
edited Nov 29 '18 at 8:58
dkaeae
304311
304311
asked Nov 28 '18 at 15:30
Dany Pépin
112
asked Nov 28 '18 at 15:30
Dany Pépin
112
asked Nov 28 '18 at 15:30
Dany Pépin
112
112
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Assume $a_n=b_n+pn+q$ therefore $$$$
therefore $$a_n=b_n+3n+2$$therefore $$a_{n}{=b_{n}+pn+q\=2b_{n-1}+3n-4+pn+q\=(3+p)n+q-4+2(a_{n-1}-pn+p-q)\=2a_{n-1}+n(3-p)+2p-4-q}$$by defining $p=3$ and $q=2$ we obtain $$a_n=2a_{n-1}$$ and since $a_0=b_0+2=4$ we conclude that $a_n=2^{n+2}$ and$$b_n=2^{n+2}-3n-2$$
answered Nov 29 '18 at 9:36
Mostafa Ayaz
13.7k3936
First, thank you for your answer, but i'm kinda unsure what made you define $a_n$ like that ? Also why did you replace it there in the $b_n$ thank you !
– Dany Pépin
Nov 29 '18 at 12:53
You're welcome. This is because a linear term like $3n-4$ strongly should have a linear source too (in some cases this is not possible). This made me think that $a_n=b_n+pn+q$ can be such that the representation of $a_{n+1}$ as a linear function of its priors, does not depend on $n$ (or any linear combination of n such as $3n-4$ in this case). By a simple calculation, I derived $p$ and $q$ and substituted. Also once you could simplify $b_n$ to $a_n$ you'd be better substitute it anywhere with $a_n$ to make your proof easier.
– Mostafa Ayaz
Nov 29 '18 at 13:00
Wait, you used derivative to obtain $a_n$ ? not sure we saw that in class haha, how would you obtain it without it ? I'm unsure. thanks !
– Dany Pépin
Nov 29 '18 at 13:44
I didn't say that! Derivation is not allowed in concrete mathematics.....
– Mostafa Ayaz
Nov 29 '18 at 13:46
Then i'm not quite sure how you landed on $b_n$ $+3n +2$ hahaha
– Dany Pépin
Nov 29 '18 at 14:14
|
show 3 more comments
Let $b_m=a_m+p+qm$
$3n-4=a_n-2a_{n-1}+p+qn-2(p+(q-1)n)$
Set $p-2p=-4$ and $2-q=3$ to find $a_n=2a_{n-1}=2^ra_{n-r}$
answered Nov 28 '18 at 15:47
lab bhattacharjee
223k15156274
What i've done : $2(1-2^{n+1})/(1-2)$ which result in $2^{n+2}-2$ seems like i have the same answer as you but from wolframalpha, the answer should be : $b_n =-3n +2^{n+2} - 2$
– Dany Pépin
Nov 28 '18 at 16:06
add a comment |
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2 Answers
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2 Answers
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Assume $a_n=b_n+pn+q$ therefore $$$$
therefore $$a_n=b_n+3n+2$$therefore $$a_{n}{=b_{n}+pn+q\=2b_{n-1}+3n-4+pn+q\=(3+p)n+q-4+2(a_{n-1}-pn+p-q)\=2a_{n-1}+n(3-p)+2p-4-q}$$by defining $p=3$ and $q=2$ we obtain $$a_n=2a_{n-1}$$ and since $a_0=b_0+2=4$ we conclude that $a_n=2^{n+2}$ and$$b_n=2^{n+2}-3n-2$$
answered Nov 29 '18 at 9:36
Mostafa Ayaz
13.7k3936
First, thank you for your answer, but i'm kinda unsure what made you define $a_n$ like that ? Also why did you replace it there in the $b_n$ thank you !
– Dany Pépin
Nov 29 '18 at 12:53
You're welcome. This is because a linear term like $3n-4$ strongly should have a linear source too (in some cases this is not possible). This made me think that $a_n=b_n+pn+q$ can be such that the representation of $a_{n+1}$ as a linear function of its priors, does not depend on $n$ (or any linear combination of n such as $3n-4$ in this case). By a simple calculation, I derived $p$ and $q$ and substituted. Also once you could simplify $b_n$ to $a_n$ you'd be better substitute it anywhere with $a_n$ to make your proof easier.
– Mostafa Ayaz
Nov 29 '18 at 13:00
Wait, you used derivative to obtain $a_n$ ? not sure we saw that in class haha, how would you obtain it without it ? I'm unsure. thanks !
– Dany Pépin
Nov 29 '18 at 13:44
I didn't say that! Derivation is not allowed in concrete mathematics.....
– Mostafa Ayaz
Nov 29 '18 at 13:46
Then i'm not quite sure how you landed on $b_n$ $+3n +2$ hahaha
– Dany Pépin
Nov 29 '18 at 14:14
|
show 3 more comments
Assume $a_n=b_n+pn+q$ therefore $$$$
therefore $$a_n=b_n+3n+2$$therefore $$a_{n}{=b_{n}+pn+q\=2b_{n-1}+3n-4+pn+q\=(3+p)n+q-4+2(a_{n-1}-pn+p-q)\=2a_{n-1}+n(3-p)+2p-4-q}$$by defining $p=3$ and $q=2$ we obtain $$a_n=2a_{n-1}$$ and since $a_0=b_0+2=4$ we conclude that $a_n=2^{n+2}$ and$$b_n=2^{n+2}-3n-2$$
answered Nov 29 '18 at 9:36
Mostafa Ayaz
13.7k3936
First, thank you for your answer, but i'm kinda unsure what made you define $a_n$ like that ? Also why did you replace it there in the $b_n$ thank you !
– Dany Pépin
Nov 29 '18 at 12:53
You're welcome. This is because a linear term like $3n-4$ strongly should have a linear source too (in some cases this is not possible). This made me think that $a_n=b_n+pn+q$ can be such that the representation of $a_{n+1}$ as a linear function of its priors, does not depend on $n$ (or any linear combination of n such as $3n-4$ in this case). By a simple calculation, I derived $p$ and $q$ and substituted. Also once you could simplify $b_n$ to $a_n$ you'd be better substitute it anywhere with $a_n$ to make your proof easier.
– Mostafa Ayaz
Nov 29 '18 at 13:00
Wait, you used derivative to obtain $a_n$ ? not sure we saw that in class haha, how would you obtain it without it ? I'm unsure. thanks !
– Dany Pépin
Nov 29 '18 at 13:44
I didn't say that! Derivation is not allowed in concrete mathematics.....
– Mostafa Ayaz
Nov 29 '18 at 13:46
Then i'm not quite sure how you landed on $b_n$ $+3n +2$ hahaha
– Dany Pépin
Nov 29 '18 at 14:14
|
show 3 more comments
Assume $a_n=b_n+pn+q$ therefore $$$$
therefore $$a_n=b_n+3n+2$$therefore $$a_{n}{=b_{n}+pn+q\=2b_{n-1}+3n-4+pn+q\=(3+p)n+q-4+2(a_{n-1}-pn+p-q)\=2a_{n-1}+n(3-p)+2p-4-q}$$by defining $p=3$ and $q=2$ we obtain $$a_n=2a_{n-1}$$ and since $a_0=b_0+2=4$ we conclude that $a_n=2^{n+2}$ and$$b_n=2^{n+2}-3n-2$$
answered Nov 29 '18 at 9:36
Mostafa Ayaz
13.7k3936
Assume $a_n=b_n+pn+q$ therefore $$$$
therefore $$a_n=b_n+3n+2$$therefore $$a_{n}{=b_{n}+pn+q\=2b_{n-1}+3n-4+pn+q\=(3+p)n+q-4+2(a_{n-1}-pn+p-q)\=2a_{n-1}+n(3-p)+2p-4-q}$$by defining $p=3$ and $q=2$ we obtain $$a_n=2a_{n-1}$$ and since $a_0=b_0+2=4$ we conclude that $a_n=2^{n+2}$ and$$b_n=2^{n+2}-3n-2$$
answered Nov 29 '18 at 9:36
Mostafa Ayaz
13.7k3936
edited Nov 29 '18 at 14:20
answered Nov 29 '18 at 9:36
Mostafa Ayaz
13.7k3936
answered Nov 29 '18 at 9:36
Mostafa Ayaz
13.7k3936
answered Nov 29 '18 at 9:36
Mostafa Ayaz
13.7k3936
13.7k3936
First, thank you for your answer, but i'm kinda unsure what made you define $a_n$ like that ? Also why did you replace it there in the $b_n$ thank you !
– Dany Pépin
Nov 29 '18 at 12:53
You're welcome. This is because a linear term like $3n-4$ strongly should have a linear source too (in some cases this is not possible). This made me think that $a_n=b_n+pn+q$ can be such that the representation of $a_{n+1}$ as a linear function of its priors, does not depend on $n$ (or any linear combination of n such as $3n-4$ in this case). By a simple calculation, I derived $p$ and $q$ and substituted. Also once you could simplify $b_n$ to $a_n$ you'd be better substitute it anywhere with $a_n$ to make your proof easier.
– Mostafa Ayaz
Nov 29 '18 at 13:00
Wait, you used derivative to obtain $a_n$ ? not sure we saw that in class haha, how would you obtain it without it ? I'm unsure. thanks !
– Dany Pépin
Nov 29 '18 at 13:44
I didn't say that! Derivation is not allowed in concrete mathematics.....
– Mostafa Ayaz
Nov 29 '18 at 13:46
Then i'm not quite sure how you landed on $b_n$ $+3n +2$ hahaha
– Dany Pépin
Nov 29 '18 at 14:14
|
show 3 more comments
First, thank you for your answer, but i'm kinda unsure what made you define $a_n$ like that ? Also why did you replace it there in the $b_n$ thank you !
– Dany Pépin
Nov 29 '18 at 12:53
You're welcome. This is because a linear term like $3n-4$ strongly should have a linear source too (in some cases this is not possible). This made me think that $a_n=b_n+pn+q$ can be such that the representation of $a_{n+1}$ as a linear function of its priors, does not depend on $n$ (or any linear combination of n such as $3n-4$ in this case). By a simple calculation, I derived $p$ and $q$ and substituted. Also once you could simplify $b_n$ to $a_n$ you'd be better substitute it anywhere with $a_n$ to make your proof easier.
– Mostafa Ayaz
Nov 29 '18 at 13:00
Wait, you used derivative to obtain $a_n$ ? not sure we saw that in class haha, how would you obtain it without it ? I'm unsure. thanks !
– Dany Pépin
Nov 29 '18 at 13:44
I didn't say that! Derivation is not allowed in concrete mathematics.....
– Mostafa Ayaz
Nov 29 '18 at 13:46
Then i'm not quite sure how you landed on $b_n$ $+3n +2$ hahaha
– Dany Pépin
Nov 29 '18 at 14:14
First, thank you for your answer, but i'm kinda unsure what made you define $a_n$ like that ? Also why did you replace it there in the $b_n$ thank you !
– Dany Pépin
Nov 29 '18 at 12:53
First, thank you for your answer, but i'm kinda unsure what made you define $a_n$ like that ? Also why did you replace it there in the $b_n$ thank you !
– Dany Pépin
Nov 29 '18 at 12:53
You're welcome. This is because a linear term like $3n-4$ strongly should have a linear source too (in some cases this is not possible). This made me think that $a_n=b_n+pn+q$ can be such that the representation of $a_{n+1}$ as a linear function of its priors, does not depend on $n$ (or any linear combination of n such as $3n-4$ in this case). By a simple calculation, I derived $p$ and $q$ and substituted. Also once you could simplify $b_n$ to $a_n$ you'd be better substitute it anywhere with $a_n$ to make your proof easier.
– Mostafa Ayaz
Nov 29 '18 at 13:00
You're welcome. This is because a linear term like $3n-4$ strongly should have a linear source too (in some cases this is not possible). This made me think that $a_n=b_n+pn+q$ can be such that the representation of $a_{n+1}$ as a linear function of its priors, does not depend on $n$ (or any linear combination of n such as $3n-4$ in this case). By a simple calculation, I derived $p$ and $q$ and substituted. Also once you could simplify $b_n$ to $a_n$ you'd be better substitute it anywhere with $a_n$ to make your proof easier.
– Mostafa Ayaz
Nov 29 '18 at 13:00
Wait, you used derivative to obtain $a_n$ ? not sure we saw that in class haha, how would you obtain it without it ? I'm unsure. thanks !
– Dany Pépin
Nov 29 '18 at 13:44
Wait, you used derivative to obtain $a_n$ ? not sure we saw that in class haha, how would you obtain it without it ? I'm unsure. thanks !
– Dany Pépin
Nov 29 '18 at 13:44
I didn't say that! Derivation is not allowed in concrete mathematics.....
– Mostafa Ayaz
Nov 29 '18 at 13:46
I didn't say that! Derivation is not allowed in concrete mathematics.....
– Mostafa Ayaz
Nov 29 '18 at 13:46
Then i'm not quite sure how you landed on $b_n$ $+3n +2$ hahaha
– Dany Pépin
Nov 29 '18 at 14:14
Then i'm not quite sure how you landed on $b_n$ $+3n +2$ hahaha
– Dany Pépin
Nov 29 '18 at 14:14
|
show 3 more comments
Let $b_m=a_m+p+qm$
$3n-4=a_n-2a_{n-1}+p+qn-2(p+(q-1)n)$
Set $p-2p=-4$ and $2-q=3$ to find $a_n=2a_{n-1}=2^ra_{n-r}$
answered Nov 28 '18 at 15:47
lab bhattacharjee
223k15156274
What i've done : $2(1-2^{n+1})/(1-2)$ which result in $2^{n+2}-2$ seems like i have the same answer as you but from wolframalpha, the answer should be : $b_n =-3n +2^{n+2} - 2$
– Dany Pépin
Nov 28 '18 at 16:06
add a comment |
Let $b_m=a_m+p+qm$
$3n-4=a_n-2a_{n-1}+p+qn-2(p+(q-1)n)$
Set $p-2p=-4$ and $2-q=3$ to find $a_n=2a_{n-1}=2^ra_{n-r}$
answered Nov 28 '18 at 15:47
lab bhattacharjee
223k15156274
What i've done : $2(1-2^{n+1})/(1-2)$ which result in $2^{n+2}-2$ seems like i have the same answer as you but from wolframalpha, the answer should be : $b_n =-3n +2^{n+2} - 2$
– Dany Pépin
Nov 28 '18 at 16:06
add a comment |
Let $b_m=a_m+p+qm$
$3n-4=a_n-2a_{n-1}+p+qn-2(p+(q-1)n)$
Set $p-2p=-4$ and $2-q=3$ to find $a_n=2a_{n-1}=2^ra_{n-r}$
answered Nov 28 '18 at 15:47
lab bhattacharjee
223k15156274
Let $b_m=a_m+p+qm$
$3n-4=a_n-2a_{n-1}+p+qn-2(p+(q-1)n)$
Set $p-2p=-4$ and $2-q=3$ to find $a_n=2a_{n-1}=2^ra_{n-r}$
answered Nov 28 '18 at 15:47
lab bhattacharjee
223k15156274
answered Nov 28 '18 at 15:47
lab bhattacharjee
223k15156274
answered Nov 28 '18 at 15:47
lab bhattacharjee
223k15156274
answered Nov 28 '18 at 15:47
lab bhattacharjee
223k15156274
223k15156274
What i've done : $2(1-2^{n+1})/(1-2)$ which result in $2^{n+2}-2$ seems like i have the same answer as you but from wolframalpha, the answer should be : $b_n =-3n +2^{n+2} - 2$
– Dany Pépin
Nov 28 '18 at 16:06
add a comment |
What i've done : $2(1-2^{n+1})/(1-2)$ which result in $2^{n+2}-2$ seems like i have the same answer as you but from wolframalpha, the answer should be : $b_n =-3n +2^{n+2} - 2$
– Dany Pépin
Nov 28 '18 at 16:06
What i've done : $2(1-2^{n+1})/(1-2)$ which result in $2^{n+2}-2$ seems like i have the same answer as you but from wolframalpha, the answer should be : $b_n =-3n +2^{n+2} - 2$
– Dany Pépin
Nov 28 '18 at 16:06
What i've done : $2(1-2^{n+1})/(1-2)$ which result in $2^{n+2}-2$ seems like i have the same answer as you but from wolframalpha, the answer should be : $b_n =-3n +2^{n+2} - 2$
– Dany Pépin
Nov 28 '18 at 16:06
add a comment |
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