Mixing
Solve recursion $a_n=a_{n-1}-6cdot3^{n-1}$ for $n>0, a_0=0$
$begingroup$
$a_n=a_{n-1}-6cdot3^{n-1}$ for $n>0, a_0=0$
So I calculate first terms
$a_0=0$
$a_1=-6$
$a_2=-24$
$a_3=-78$
I don't see any relation so
$a_n=a_{n-1}-6cdot3^{n-1}$
$a_{n-1}=a_{n-2}-6cdot 3^{n-2}$
. . .
$a_2=a_1-6cdot3^{1}$
$a_1=a_0-6cdot 3^{0}$
Not sure what to do next, Wolfram solves it in this way:
$a_n=-3cdot(3^{n}-1)$
How do I get to this point?
discrete-mathematics
edited Jan 27 at 13:13
Stefan4024
30.6k63579
asked Jan 27 at 13:06
GorossoGorosso
315
$endgroup$
add a comment |
$begingroup$
$a_n=a_{n-1}-6cdot3^{n-1}$ for $n>0, a_0=0$
So I calculate first terms
$a_0=0$
$a_1=-6$
$a_2=-24$
$a_3=-78$
I don't see any relation so
$a_n=a_{n-1}-6cdot3^{n-1}$
$a_{n-1}=a_{n-2}-6cdot 3^{n-2}$
. . .
$a_2=a_1-6cdot3^{1}$
$a_1=a_0-6cdot 3^{0}$
Not sure what to do next, Wolfram solves it in this way:
$a_n=-3cdot(3^{n}-1)$
How do I get to this point?
discrete-mathematics
edited Jan 27 at 13:13
Stefan4024
30.6k63579
asked Jan 27 at 13:06
GorossoGorosso
315
$endgroup$
$begingroup$
Welcome to TeX SX! Do you mean $a_n$ or $an$?
$endgroup$
– Bernard
Jan 27 at 13:07
$begingroup$
Hello, yes I meant first option, didn't know how to do it in tex, sorry.
$endgroup$
– Gorosso
Jan 27 at 13:14
$begingroup$
it's justa_{n}, a_{n-1}, &c. If there's only 1 character in subscript, the braces are not necessary.
$endgroup$
– Bernard
Jan 27 at 13:21
add a comment |
$begingroup$
$a_n=a_{n-1}-6cdot3^{n-1}$ for $n>0, a_0=0$
So I calculate first terms
$a_0=0$
$a_1=-6$
$a_2=-24$
$a_3=-78$
I don't see any relation so
$a_n=a_{n-1}-6cdot3^{n-1}$
$a_{n-1}=a_{n-2}-6cdot 3^{n-2}$
. . .
$a_2=a_1-6cdot3^{1}$
$a_1=a_0-6cdot 3^{0}$
Not sure what to do next, Wolfram solves it in this way:
$a_n=-3cdot(3^{n}-1)$
How do I get to this point?
discrete-mathematics
edited Jan 27 at 13:13
Stefan4024
30.6k63579
asked Jan 27 at 13:06
GorossoGorosso
315
$endgroup$
$a_n=a_{n-1}-6cdot3^{n-1}$ for $n>0, a_0=0$
So I calculate first terms
$a_0=0$
$a_1=-6$
$a_2=-24$
$a_3=-78$
I don't see any relation so
$a_n=a_{n-1}-6cdot3^{n-1}$
$a_{n-1}=a_{n-2}-6cdot 3^{n-2}$
. . .
$a_2=a_1-6cdot3^{1}$
$a_1=a_0-6cdot 3^{0}$
Not sure what to do next, Wolfram solves it in this way:
$a_n=-3cdot(3^{n}-1)$
How do I get to this point?
discrete-mathematics
discrete-mathematics
edited Jan 27 at 13:13
Stefan4024
30.6k63579
asked Jan 27 at 13:06
GorossoGorosso
315
edited Jan 27 at 13:13
Stefan4024
30.6k63579
asked Jan 27 at 13:06
GorossoGorosso
315
edited Jan 27 at 13:13
Stefan4024
30.6k63579
edited Jan 27 at 13:13
Stefan4024
30.6k63579
edited Jan 27 at 13:13
Stefan4024
30.6k63579
30.6k63579
asked Jan 27 at 13:06
GorossoGorosso
315
asked Jan 27 at 13:06
GorossoGorosso
315
asked Jan 27 at 13:06
GorossoGorosso
315
315
$begingroup$
Welcome to TeX SX! Do you mean $a_n$ or $an$?
$endgroup$
– Bernard
Jan 27 at 13:07
$begingroup$
Hello, yes I meant first option, didn't know how to do it in tex, sorry.
$endgroup$
– Gorosso
Jan 27 at 13:14
$begingroup$
it's justa_{n}, a_{n-1}, &c. If there's only 1 character in subscript, the braces are not necessary.
$endgroup$
– Bernard
Jan 27 at 13:21
add a comment |
$begingroup$
Welcome to TeX SX! Do you mean $a_n$ or $an$?
$endgroup$
– Bernard
Jan 27 at 13:07
$begingroup$
Hello, yes I meant first option, didn't know how to do it in tex, sorry.
$endgroup$
– Gorosso
Jan 27 at 13:14
$begingroup$
it's justa_{n}, a_{n-1}, &c. If there's only 1 character in subscript, the braces are not necessary.
$endgroup$
– Bernard
Jan 27 at 13:21
$begingroup$
Welcome to TeX SX! Do you mean $a_n$ or $an$?
$endgroup$
– Bernard
Jan 27 at 13:07
$begingroup$
Welcome to TeX SX! Do you mean $a_n$ or $an$?
$endgroup$
– Bernard
Jan 27 at 13:07
$begingroup$
Hello, yes I meant first option, didn't know how to do it in tex, sorry.
$endgroup$
– Gorosso
Jan 27 at 13:14
$begingroup$
Hello, yes I meant first option, didn't know how to do it in tex, sorry.
$endgroup$
– Gorosso
Jan 27 at 13:14
$begingroup$
it's just
a_{n}, a_{n-1}, &c. If there's only 1 character in subscript, the braces are not necessary.$endgroup$
– Bernard
Jan 27 at 13:21
$begingroup$
it's just
a_{n}, a_{n-1}, &c. If there's only 1 character in subscript, the braces are not necessary.$endgroup$
– Bernard
Jan 27 at 13:21
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You have
$$a_n=a_n=a_{n-1}-6cdot 3^{n-1}=a_{n-2}-6cdot 3^{n-2}-6cdot 3^{n-1}=dotsm,$$
so you can prove with an easy induction that
$$a_n = a_0-6sum_{k=0}^{n-1} 3^k=-6frac{3^n-1}{3-1}.$$
answered Jan 27 at 13:19
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
hint...consider $$sum_{r=0}^{r=n}(a_{r+1}-a_r)=sum_{r=0}^{r=n}-6cdot3^r$$
The LHS is a telescoping series and the RHS is a geometric series.
answered Jan 27 at 13:20
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
This is just a Geometric Series:
$$a_n=-6sum_{i=1}^n3^{i-1}=-6sum_{i=0}^{n-1}3^i=-6times frac {3^n-1}{3-1}=-3times (3^n-1)$$
answered Jan 27 at 13:14
lulululu
43.2k25080
$endgroup$
add a comment |
$begingroup$
Note that for all $n$ we have $$a_{n+1}-a_n = -6cdot 3^n $$
so we have also: $$a_n-a_{n-1}=-6cdot3^{n-1}$$
thus $$a_{n+1}-a_n = 3(-6cdot3^{n-1}) = 3(a_n-a_{n-1})$$
{or divide this two equations: $${a_{n+1}-a_n over a_n-a_{n-1}}= {-6cdot 3^nover -6cdot3^{n-1}} = 3$$}
so you have to solve linear recurrence:
$$ a_{n+1}-4a_n+3a_{n-1}=0$$
CAn you do that?
answered Jan 27 at 13:12
Maria MazurMaria Mazur
48.3k1260121
$endgroup$
$begingroup$
ok so: $x^{2}-4x+3$ $Delta =4$ $x1=1$ $x2=3$ Now I use following formula $a_n=ar^{n}+br^{n}$ $a_0=0=a+b$ $a_1=-6=a+3b$ $a=-b$ $-6=2b$ $b=-3$ $a=3$ $a_n=3*1^{n}-3*3^{n}$
$endgroup$
– Gorosso
Jan 27 at 14:40
1
$begingroup$
Yes, that is correct!
$endgroup$
– Maria Mazur
Jan 27 at 14:40
1
$begingroup$
Is that better?
$endgroup$
– Maria Mazur
Jan 27 at 14:52
1
$begingroup$
What about now?
$endgroup$
– Maria Mazur
Jan 27 at 15:51
1
$begingroup$
Just put first and last expression on one side, so you get 0 on other side
$endgroup$
– Maria Mazur
Jan 27 at 16:54
|
show 4 more comments
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have
$$a_n=a_n=a_{n-1}-6cdot 3^{n-1}=a_{n-2}-6cdot 3^{n-2}-6cdot 3^{n-1}=dotsm,$$
so you can prove with an easy induction that
$$a_n = a_0-6sum_{k=0}^{n-1} 3^k=-6frac{3^n-1}{3-1}.$$
answered Jan 27 at 13:19
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
You have
$$a_n=a_n=a_{n-1}-6cdot 3^{n-1}=a_{n-2}-6cdot 3^{n-2}-6cdot 3^{n-1}=dotsm,$$
so you can prove with an easy induction that
$$a_n = a_0-6sum_{k=0}^{n-1} 3^k=-6frac{3^n-1}{3-1}.$$
answered Jan 27 at 13:19
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
You have
$$a_n=a_n=a_{n-1}-6cdot 3^{n-1}=a_{n-2}-6cdot 3^{n-2}-6cdot 3^{n-1}=dotsm,$$
so you can prove with an easy induction that
$$a_n = a_0-6sum_{k=0}^{n-1} 3^k=-6frac{3^n-1}{3-1}.$$
answered Jan 27 at 13:19
BernardBernard
123k741117
$endgroup$
You have
$$a_n=a_n=a_{n-1}-6cdot 3^{n-1}=a_{n-2}-6cdot 3^{n-2}-6cdot 3^{n-1}=dotsm,$$
so you can prove with an easy induction that
$$a_n = a_0-6sum_{k=0}^{n-1} 3^k=-6frac{3^n-1}{3-1}.$$
answered Jan 27 at 13:19
BernardBernard
123k741117
answered Jan 27 at 13:19
BernardBernard
123k741117
answered Jan 27 at 13:19
BernardBernard
123k741117
answered Jan 27 at 13:19
BernardBernard
123k741117
123k741117
add a comment |
add a comment |
$begingroup$
hint...consider $$sum_{r=0}^{r=n}(a_{r+1}-a_r)=sum_{r=0}^{r=n}-6cdot3^r$$
The LHS is a telescoping series and the RHS is a geometric series.
answered Jan 27 at 13:20
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
hint...consider $$sum_{r=0}^{r=n}(a_{r+1}-a_r)=sum_{r=0}^{r=n}-6cdot3^r$$
The LHS is a telescoping series and the RHS is a geometric series.
answered Jan 27 at 13:20
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
hint...consider $$sum_{r=0}^{r=n}(a_{r+1}-a_r)=sum_{r=0}^{r=n}-6cdot3^r$$
The LHS is a telescoping series and the RHS is a geometric series.
answered Jan 27 at 13:20
David QuinnDavid Quinn
24.1k21141
$endgroup$
hint...consider $$sum_{r=0}^{r=n}(a_{r+1}-a_r)=sum_{r=0}^{r=n}-6cdot3^r$$
The LHS is a telescoping series and the RHS is a geometric series.
answered Jan 27 at 13:20
David QuinnDavid Quinn
24.1k21141
answered Jan 27 at 13:20
David QuinnDavid Quinn
24.1k21141
answered Jan 27 at 13:20
David QuinnDavid Quinn
24.1k21141
answered Jan 27 at 13:20
David QuinnDavid Quinn
24.1k21141
24.1k21141
add a comment |
add a comment |
$begingroup$
This is just a Geometric Series:
$$a_n=-6sum_{i=1}^n3^{i-1}=-6sum_{i=0}^{n-1}3^i=-6times frac {3^n-1}{3-1}=-3times (3^n-1)$$
answered Jan 27 at 13:14
lulululu
43.2k25080
$endgroup$
add a comment |
$begingroup$
This is just a Geometric Series:
$$a_n=-6sum_{i=1}^n3^{i-1}=-6sum_{i=0}^{n-1}3^i=-6times frac {3^n-1}{3-1}=-3times (3^n-1)$$
answered Jan 27 at 13:14
lulululu
43.2k25080
$endgroup$
add a comment |
$begingroup$
This is just a Geometric Series:
$$a_n=-6sum_{i=1}^n3^{i-1}=-6sum_{i=0}^{n-1}3^i=-6times frac {3^n-1}{3-1}=-3times (3^n-1)$$
answered Jan 27 at 13:14
lulululu
43.2k25080
$endgroup$
This is just a Geometric Series:
$$a_n=-6sum_{i=1}^n3^{i-1}=-6sum_{i=0}^{n-1}3^i=-6times frac {3^n-1}{3-1}=-3times (3^n-1)$$
answered Jan 27 at 13:14
lulululu
43.2k25080
answered Jan 27 at 13:14
lulululu
43.2k25080
answered Jan 27 at 13:14
lulululu
43.2k25080
answered Jan 27 at 13:14
lulululu
43.2k25080
43.2k25080
add a comment |
add a comment |
$begingroup$
Note that for all $n$ we have $$a_{n+1}-a_n = -6cdot 3^n $$
so we have also: $$a_n-a_{n-1}=-6cdot3^{n-1}$$
thus $$a_{n+1}-a_n = 3(-6cdot3^{n-1}) = 3(a_n-a_{n-1})$$
{or divide this two equations: $${a_{n+1}-a_n over a_n-a_{n-1}}= {-6cdot 3^nover -6cdot3^{n-1}} = 3$$}
so you have to solve linear recurrence:
$$ a_{n+1}-4a_n+3a_{n-1}=0$$
CAn you do that?
answered Jan 27 at 13:12
Maria MazurMaria Mazur
48.3k1260121
$endgroup$
$begingroup$
ok so: $x^{2}-4x+3$ $Delta =4$ $x1=1$ $x2=3$ Now I use following formula $a_n=ar^{n}+br^{n}$ $a_0=0=a+b$ $a_1=-6=a+3b$ $a=-b$ $-6=2b$ $b=-3$ $a=3$ $a_n=3*1^{n}-3*3^{n}$
$endgroup$
– Gorosso
Jan 27 at 14:40
1
$begingroup$
Yes, that is correct!
$endgroup$
– Maria Mazur
Jan 27 at 14:40
1
$begingroup$
Is that better?
$endgroup$
– Maria Mazur
Jan 27 at 14:52
1
$begingroup$
What about now?
$endgroup$
– Maria Mazur
Jan 27 at 15:51
1
$begingroup$
Just put first and last expression on one side, so you get 0 on other side
$endgroup$
– Maria Mazur
Jan 27 at 16:54
|
show 4 more comments
$begingroup$
Note that for all $n$ we have $$a_{n+1}-a_n = -6cdot 3^n $$
so we have also: $$a_n-a_{n-1}=-6cdot3^{n-1}$$
thus $$a_{n+1}-a_n = 3(-6cdot3^{n-1}) = 3(a_n-a_{n-1})$$
{or divide this two equations: $${a_{n+1}-a_n over a_n-a_{n-1}}= {-6cdot 3^nover -6cdot3^{n-1}} = 3$$}
so you have to solve linear recurrence:
$$ a_{n+1}-4a_n+3a_{n-1}=0$$
CAn you do that?
answered Jan 27 at 13:12
Maria MazurMaria Mazur
48.3k1260121
$endgroup$
$begingroup$
ok so: $x^{2}-4x+3$ $Delta =4$ $x1=1$ $x2=3$ Now I use following formula $a_n=ar^{n}+br^{n}$ $a_0=0=a+b$ $a_1=-6=a+3b$ $a=-b$ $-6=2b$ $b=-3$ $a=3$ $a_n=3*1^{n}-3*3^{n}$
$endgroup$
– Gorosso
Jan 27 at 14:40
1
$begingroup$
Yes, that is correct!
$endgroup$
– Maria Mazur
Jan 27 at 14:40
1
$begingroup$
Is that better?
$endgroup$
– Maria Mazur
Jan 27 at 14:52
1
$begingroup$
What about now?
$endgroup$
– Maria Mazur
Jan 27 at 15:51
1
$begingroup$
Just put first and last expression on one side, so you get 0 on other side
$endgroup$
– Maria Mazur
Jan 27 at 16:54
|
show 4 more comments
$begingroup$
Note that for all $n$ we have $$a_{n+1}-a_n = -6cdot 3^n $$
so we have also: $$a_n-a_{n-1}=-6cdot3^{n-1}$$
thus $$a_{n+1}-a_n = 3(-6cdot3^{n-1}) = 3(a_n-a_{n-1})$$
{or divide this two equations: $${a_{n+1}-a_n over a_n-a_{n-1}}= {-6cdot 3^nover -6cdot3^{n-1}} = 3$$}
so you have to solve linear recurrence:
$$ a_{n+1}-4a_n+3a_{n-1}=0$$
CAn you do that?
answered Jan 27 at 13:12
Maria MazurMaria Mazur
48.3k1260121
$endgroup$
Note that for all $n$ we have $$a_{n+1}-a_n = -6cdot 3^n $$
so we have also: $$a_n-a_{n-1}=-6cdot3^{n-1}$$
thus $$a_{n+1}-a_n = 3(-6cdot3^{n-1}) = 3(a_n-a_{n-1})$$
{or divide this two equations: $${a_{n+1}-a_n over a_n-a_{n-1}}= {-6cdot 3^nover -6cdot3^{n-1}} = 3$$}
so you have to solve linear recurrence:
$$ a_{n+1}-4a_n+3a_{n-1}=0$$
CAn you do that?
answered Jan 27 at 13:12
Maria MazurMaria Mazur
48.3k1260121
edited Jan 27 at 15:51
answered Jan 27 at 13:12
Maria MazurMaria Mazur
48.3k1260121
answered Jan 27 at 13:12
Maria MazurMaria Mazur
48.3k1260121
answered Jan 27 at 13:12
Maria MazurMaria Mazur
48.3k1260121
48.3k1260121
$begingroup$
ok so: $x^{2}-4x+3$ $Delta =4$ $x1=1$ $x2=3$ Now I use following formula $a_n=ar^{n}+br^{n}$ $a_0=0=a+b$ $a_1=-6=a+3b$ $a=-b$ $-6=2b$ $b=-3$ $a=3$ $a_n=3*1^{n}-3*3^{n}$
$endgroup$
– Gorosso
Jan 27 at 14:40
1
$begingroup$
Yes, that is correct!
$endgroup$
– Maria Mazur
Jan 27 at 14:40
1
$begingroup$
Is that better?
$endgroup$
– Maria Mazur
Jan 27 at 14:52
1
$begingroup$
What about now?
$endgroup$
– Maria Mazur
Jan 27 at 15:51
1
$begingroup$
Just put first and last expression on one side, so you get 0 on other side
$endgroup$
– Maria Mazur
Jan 27 at 16:54
|
show 4 more comments
$begingroup$
ok so: $x^{2}-4x+3$ $Delta =4$ $x1=1$ $x2=3$ Now I use following formula $a_n=ar^{n}+br^{n}$ $a_0=0=a+b$ $a_1=-6=a+3b$ $a=-b$ $-6=2b$ $b=-3$ $a=3$ $a_n=3*1^{n}-3*3^{n}$
$endgroup$
– Gorosso
Jan 27 at 14:40
1
$begingroup$
Yes, that is correct!
$endgroup$
– Maria Mazur
Jan 27 at 14:40
1
$begingroup$
Is that better?
$endgroup$
– Maria Mazur
Jan 27 at 14:52
1
$begingroup$
What about now?
$endgroup$
– Maria Mazur
Jan 27 at 15:51
1
$begingroup$
Just put first and last expression on one side, so you get 0 on other side
$endgroup$
– Maria Mazur
Jan 27 at 16:54
$begingroup$
ok so: $x^{2}-4x+3$ $Delta =4$ $x1=1$ $x2=3$ Now I use following formula $a_n=ar^{n}+br^{n}$ $a_0=0=a+b$ $a_1=-6=a+3b$ $a=-b$ $-6=2b$ $b=-3$ $a=3$ $a_n=3*1^{n}-3*3^{n}$
$endgroup$
– Gorosso
Jan 27 at 14:40
$begingroup$
ok so: $x^{2}-4x+3$ $Delta =4$ $x1=1$ $x2=3$ Now I use following formula $a_n=ar^{n}+br^{n}$ $a_0=0=a+b$ $a_1=-6=a+3b$ $a=-b$ $-6=2b$ $b=-3$ $a=3$ $a_n=3*1^{n}-3*3^{n}$
$endgroup$
– Gorosso
Jan 27 at 14:40
1
1
$begingroup$
Yes, that is correct!
$endgroup$
– Maria Mazur
Jan 27 at 14:40
$begingroup$
Yes, that is correct!
$endgroup$
– Maria Mazur
Jan 27 at 14:40
1
1
$begingroup$
Is that better?
$endgroup$
– Maria Mazur
Jan 27 at 14:52
$begingroup$
Is that better?
$endgroup$
– Maria Mazur
Jan 27 at 14:52
1
1
$begingroup$
What about now?
$endgroup$
– Maria Mazur
Jan 27 at 15:51
$begingroup$
What about now?
$endgroup$
– Maria Mazur
Jan 27 at 15:51
1
1
$begingroup$
Just put first and last expression on one side, so you get 0 on other side
$endgroup$
– Maria Mazur
Jan 27 at 16:54
$begingroup$
Just put first and last expression on one side, so you get 0 on other side
$endgroup$
– Maria Mazur
Jan 27 at 16:54
|
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$begingroup$
Welcome to TeX SX! Do you mean $a_n$ or $an$?
$endgroup$
– Bernard
Jan 27 at 13:07
$begingroup$
Hello, yes I meant first option, didn't know how to do it in tex, sorry.
$endgroup$
– Gorosso
Jan 27 at 13:14
$begingroup$
it's just
a_{n}, a_{n-1}, &c. If there's only 1 character in subscript, the braces are not necessary.$endgroup$
– Bernard
Jan 27 at 13:21