Mixing
Alphabet with 6 vowels and 12 consonants, find the amount of words without two consonants in a row.
$begingroup$
I just took an exam and as usual with exams, the answers come to you when you're done with the exam and you are sitting in your favourite chair at home. I want to verify my solution as part of my learning process to learn from my mistakes in case I might want to schedule a resit
Consider an alphabet $A$ consisting of $6$ vowels and of $12$ consonants. Valid words consist of no two consonants in a row, so AART is not valid, nor is JUDITH, but JUDIT is fine and so is AAR, as is AIAIAIAIAIAIAIAIAI. $a_n$ denotes the amount of valid words.
a) find $a_0$, $a_1$, $a_2$, $a_3$
$a_0=1$, the empty word
$a_1=12+6=18$ (just one letter)
For $a_2$ we considers words like $AT$, $TA$, $IA$(different vowels) and $AA$ (same vowels)
$a_2= 2 times 6 cdot 12 + 5 cdot 6 + 6=144 +30 +6=180$
We expand to three symbols by either adding a vowel to the end of a 2-letter word or by adding a vowel and consonant to a 1-letter word
$a_3=180 cdot 6 + 6 cdot 12 cdot 18 =1080+1296=2376$
(b) Find a recurrence relation
(c) solve it
We make a case distinction for a valid word of length $n$, it either ends in a consonant or in a vowel. If it ends in a consonant, we must have obtained it from a valid word of length $n-2$ by placing a vowel followed by a consonant behind it. In all other situations we simply place a vowel behind a word of length $n-1$.
We get for $ngeq 2$:
$$ a_n = 6 cdot a_{n-1} + 6 cdot 12 cdot a_{n-2}$$
One can verify that this indeed gives $180$ for $a_2$.
We can solve this recursion via an auxiliary equation of the form:
$$ r^2 = 6r + 6 cdot 16 $$
$$ r^2 - 6r - 6 cdot 16 =0$$
Which factorises as:
$$ (r-12)(r+6)=0$$
So we get solutions $a_n = A r_1^n + B r_2^n$:
$$ a_n = A cdot 12^n + B cdot (-6) ^n$$
We can now plug in our initial conditions $a_0=1$ and $a_1=18$
$$1=A+B$$
$$ 18= 12A - 6B=18A -6 implies 18A=24 implies A=frac{4}{3}, B=-frac{1}{3}. $$
We get:
$$ a_n = frac{4}{3}cdot 12^n -frac{1}{3} (- 6)^n$$
I feel that this is probably correct, but I am unsure. Can someone please verify?
combinatorics proof-verification recurrence-relations
asked Jan 31 at 20:49
Wesley StrikWesley Strik
2,189424
$endgroup$
add a comment |
$begingroup$
I just took an exam and as usual with exams, the answers come to you when you're done with the exam and you are sitting in your favourite chair at home. I want to verify my solution as part of my learning process to learn from my mistakes in case I might want to schedule a resit
Consider an alphabet $A$ consisting of $6$ vowels and of $12$ consonants. Valid words consist of no two consonants in a row, so AART is not valid, nor is JUDITH, but JUDIT is fine and so is AAR, as is AIAIAIAIAIAIAIAIAI. $a_n$ denotes the amount of valid words.
a) find $a_0$, $a_1$, $a_2$, $a_3$
$a_0=1$, the empty word
$a_1=12+6=18$ (just one letter)
For $a_2$ we considers words like $AT$, $TA$, $IA$(different vowels) and $AA$ (same vowels)
$a_2= 2 times 6 cdot 12 + 5 cdot 6 + 6=144 +30 +6=180$
We expand to three symbols by either adding a vowel to the end of a 2-letter word or by adding a vowel and consonant to a 1-letter word
$a_3=180 cdot 6 + 6 cdot 12 cdot 18 =1080+1296=2376$
(b) Find a recurrence relation
(c) solve it
We make a case distinction for a valid word of length $n$, it either ends in a consonant or in a vowel. If it ends in a consonant, we must have obtained it from a valid word of length $n-2$ by placing a vowel followed by a consonant behind it. In all other situations we simply place a vowel behind a word of length $n-1$.
We get for $ngeq 2$:
$$ a_n = 6 cdot a_{n-1} + 6 cdot 12 cdot a_{n-2}$$
One can verify that this indeed gives $180$ for $a_2$.
We can solve this recursion via an auxiliary equation of the form:
$$ r^2 = 6r + 6 cdot 16 $$
$$ r^2 - 6r - 6 cdot 16 =0$$
Which factorises as:
$$ (r-12)(r+6)=0$$
So we get solutions $a_n = A r_1^n + B r_2^n$:
$$ a_n = A cdot 12^n + B cdot (-6) ^n$$
We can now plug in our initial conditions $a_0=1$ and $a_1=18$
$$1=A+B$$
$$ 18= 12A - 6B=18A -6 implies 18A=24 implies A=frac{4}{3}, B=-frac{1}{3}. $$
We get:
$$ a_n = frac{4}{3}cdot 12^n -frac{1}{3} (- 6)^n$$
I feel that this is probably correct, but I am unsure. Can someone please verify?
combinatorics proof-verification recurrence-relations
asked Jan 31 at 20:49
Wesley StrikWesley Strik
2,189424
$endgroup$
3
$begingroup$
Your reasoning looks fine to me :-)
$endgroup$
– Nicolas FRANCOIS
Jan 31 at 20:54
$begingroup$
You have a typo; you wrote "$r^2=6r-6cdot 16=0$", when I think you meant "$r^2-6r-6cdot 12=0$".
$endgroup$
– Mike Earnest
Jan 31 at 21:33
1
$begingroup$
$a_3=180 cdot 6 + 6 cdot 12 cdot 18 =1080+1296=2376$
$endgroup$
– Daniel Mathias
Feb 1 at 1:37
add a comment |
$begingroup$
I just took an exam and as usual with exams, the answers come to you when you're done with the exam and you are sitting in your favourite chair at home. I want to verify my solution as part of my learning process to learn from my mistakes in case I might want to schedule a resit
Consider an alphabet $A$ consisting of $6$ vowels and of $12$ consonants. Valid words consist of no two consonants in a row, so AART is not valid, nor is JUDITH, but JUDIT is fine and so is AAR, as is AIAIAIAIAIAIAIAIAI. $a_n$ denotes the amount of valid words.
a) find $a_0$, $a_1$, $a_2$, $a_3$
$a_0=1$, the empty word
$a_1=12+6=18$ (just one letter)
For $a_2$ we considers words like $AT$, $TA$, $IA$(different vowels) and $AA$ (same vowels)
$a_2= 2 times 6 cdot 12 + 5 cdot 6 + 6=144 +30 +6=180$
We expand to three symbols by either adding a vowel to the end of a 2-letter word or by adding a vowel and consonant to a 1-letter word
$a_3=180 cdot 6 + 6 cdot 12 cdot 18 =1080+1296=2376$
(b) Find a recurrence relation
(c) solve it
We make a case distinction for a valid word of length $n$, it either ends in a consonant or in a vowel. If it ends in a consonant, we must have obtained it from a valid word of length $n-2$ by placing a vowel followed by a consonant behind it. In all other situations we simply place a vowel behind a word of length $n-1$.
We get for $ngeq 2$:
$$ a_n = 6 cdot a_{n-1} + 6 cdot 12 cdot a_{n-2}$$
One can verify that this indeed gives $180$ for $a_2$.
We can solve this recursion via an auxiliary equation of the form:
$$ r^2 = 6r + 6 cdot 16 $$
$$ r^2 - 6r - 6 cdot 16 =0$$
Which factorises as:
$$ (r-12)(r+6)=0$$
So we get solutions $a_n = A r_1^n + B r_2^n$:
$$ a_n = A cdot 12^n + B cdot (-6) ^n$$
We can now plug in our initial conditions $a_0=1$ and $a_1=18$
$$1=A+B$$
$$ 18= 12A - 6B=18A -6 implies 18A=24 implies A=frac{4}{3}, B=-frac{1}{3}. $$
We get:
$$ a_n = frac{4}{3}cdot 12^n -frac{1}{3} (- 6)^n$$
I feel that this is probably correct, but I am unsure. Can someone please verify?
combinatorics proof-verification recurrence-relations
asked Jan 31 at 20:49
Wesley StrikWesley Strik
2,189424
$endgroup$
I just took an exam and as usual with exams, the answers come to you when you're done with the exam and you are sitting in your favourite chair at home. I want to verify my solution as part of my learning process to learn from my mistakes in case I might want to schedule a resit
Consider an alphabet $A$ consisting of $6$ vowels and of $12$ consonants. Valid words consist of no two consonants in a row, so AART is not valid, nor is JUDITH, but JUDIT is fine and so is AAR, as is AIAIAIAIAIAIAIAIAI. $a_n$ denotes the amount of valid words.
a) find $a_0$, $a_1$, $a_2$, $a_3$
$a_0=1$, the empty word
$a_1=12+6=18$ (just one letter)
For $a_2$ we considers words like $AT$, $TA$, $IA$(different vowels) and $AA$ (same vowels)
$a_2= 2 times 6 cdot 12 + 5 cdot 6 + 6=144 +30 +6=180$
We expand to three symbols by either adding a vowel to the end of a 2-letter word or by adding a vowel and consonant to a 1-letter word
$a_3=180 cdot 6 + 6 cdot 12 cdot 18 =1080+1296=2376$
(b) Find a recurrence relation
(c) solve it
We make a case distinction for a valid word of length $n$, it either ends in a consonant or in a vowel. If it ends in a consonant, we must have obtained it from a valid word of length $n-2$ by placing a vowel followed by a consonant behind it. In all other situations we simply place a vowel behind a word of length $n-1$.
We get for $ngeq 2$:
$$ a_n = 6 cdot a_{n-1} + 6 cdot 12 cdot a_{n-2}$$
One can verify that this indeed gives $180$ for $a_2$.
We can solve this recursion via an auxiliary equation of the form:
$$ r^2 = 6r + 6 cdot 16 $$
$$ r^2 - 6r - 6 cdot 16 =0$$
Which factorises as:
$$ (r-12)(r+6)=0$$
So we get solutions $a_n = A r_1^n + B r_2^n$:
$$ a_n = A cdot 12^n + B cdot (-6) ^n$$
We can now plug in our initial conditions $a_0=1$ and $a_1=18$
$$1=A+B$$
$$ 18= 12A - 6B=18A -6 implies 18A=24 implies A=frac{4}{3}, B=-frac{1}{3}. $$
We get:
$$ a_n = frac{4}{3}cdot 12^n -frac{1}{3} (- 6)^n$$
I feel that this is probably correct, but I am unsure. Can someone please verify?
combinatorics proof-verification recurrence-relations
combinatorics proof-verification recurrence-relations
asked Jan 31 at 20:49
Wesley StrikWesley Strik
2,189424
asked Jan 31 at 20:49
Wesley StrikWesley Strik
2,189424
edited Feb 1 at 13:00
Wesley Strik
asked Jan 31 at 20:49
Wesley StrikWesley Strik
2,189424
asked Jan 31 at 20:49
Wesley StrikWesley Strik
2,189424
asked Jan 31 at 20:49
Wesley StrikWesley Strik
2,189424
2,189424
3
$begingroup$
Your reasoning looks fine to me :-)
$endgroup$
– Nicolas FRANCOIS
Jan 31 at 20:54
$begingroup$
You have a typo; you wrote "$r^2=6r-6cdot 16=0$", when I think you meant "$r^2-6r-6cdot 12=0$".
$endgroup$
– Mike Earnest
Jan 31 at 21:33
1
$begingroup$
$a_3=180 cdot 6 + 6 cdot 12 cdot 18 =1080+1296=2376$
$endgroup$
– Daniel Mathias
Feb 1 at 1:37
add a comment |
3
$begingroup$
Your reasoning looks fine to me :-)
$endgroup$
– Nicolas FRANCOIS
Jan 31 at 20:54
$begingroup$
You have a typo; you wrote "$r^2=6r-6cdot 16=0$", when I think you meant "$r^2-6r-6cdot 12=0$".
$endgroup$
– Mike Earnest
Jan 31 at 21:33
1
$begingroup$
$a_3=180 cdot 6 + 6 cdot 12 cdot 18 =1080+1296=2376$
$endgroup$
– Daniel Mathias
Feb 1 at 1:37
3
3
$begingroup$
Your reasoning looks fine to me :-)
$endgroup$
– Nicolas FRANCOIS
Jan 31 at 20:54
$begingroup$
Your reasoning looks fine to me :-)
$endgroup$
– Nicolas FRANCOIS
Jan 31 at 20:54
$begingroup$
You have a typo; you wrote "$r^2=6r-6cdot 16=0$", when I think you meant "$r^2-6r-6cdot 12=0$".
$endgroup$
– Mike Earnest
Jan 31 at 21:33
$begingroup$
You have a typo; you wrote "$r^2=6r-6cdot 16=0$", when I think you meant "$r^2-6r-6cdot 12=0$".
$endgroup$
– Mike Earnest
Jan 31 at 21:33
1
1
$begingroup$
$a_3=180 cdot 6 + 6 cdot 12 cdot 18 =1080+1296=2376$
$endgroup$
– Daniel Mathias
Feb 1 at 1:37
$begingroup$
$a_3=180 cdot 6 + 6 cdot 12 cdot 18 =1080+1296=2376$
$endgroup$
– Daniel Mathias
Feb 1 at 1:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$a_n=6a_{n-1}+6cdot 12a_{n-2}$$ is correct. From this you get the characteristic polynomial
$$
r^2-6r-6cdot 12
$$
which factors as
$$
(r-12)(r+6)
$$
Note that the roots of this are $r=12$ and $r=-6$. This is where you went wrong; you had the roots as $r=12$ and $r=6$. The general solution is therefore
$$
a_n=Acdot 12^n+B(-6)^n
$$
and you can do the rest.
answered Jan 31 at 21:12
Mike EarnestMike Earnest
27.3k22152
$endgroup$
$begingroup$
Thanks for checking! :)
$endgroup$
– Wesley Strik
Feb 1 at 13:08
add a comment |
$begingroup$
if there are $a_n$ n-letter words, $c_n$ end in a consonant and $v_n$ end in a vowel
$a_n = c_n + v_n$
To make a n+1 letter word, we take the a_n letter words and tack a letter onto the end.
$a_n = 6c_n + 12 v_n\
c_{n+1} = 12v_n\
v_{n+1} = 6 c_n+ 6v_n = 6 a_n$
We probably could do away with $a_n$ and represent just with $c_n, v_n$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = begin{bmatrix}0&12\6 & 6end{bmatrix}begin{bmatrix} c_{n}\v_{n} end {bmatrix}$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = begin{bmatrix}0&12\6 & 6end{bmatrix}^nbegin{bmatrix} c_{1}\v_{1} end {bmatrix}$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = frac 13 begin{bmatrix}1&-2\1 & 1end{bmatrix}begin{bmatrix}12^n&0\ 0& -6^nend{bmatrix}begin{bmatrix}1&2\-1 & 1end{bmatrix}begin{bmatrix} 12\6 end {bmatrix}$
$a_n = frac {4(12)^n - (-6)^n}{3}$
answered Jan 31 at 21:51
Doug MDoug M
45.4k31954
$endgroup$
$begingroup$
This is a nice method that I will try to use myself more often :)
$endgroup$
– Wesley Strik
Feb 1 at 13:09
add a comment |
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2 Answers
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2 Answers
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votes
$begingroup$
$$a_n=6a_{n-1}+6cdot 12a_{n-2}$$ is correct. From this you get the characteristic polynomial
$$
r^2-6r-6cdot 12
$$
which factors as
$$
(r-12)(r+6)
$$
Note that the roots of this are $r=12$ and $r=-6$. This is where you went wrong; you had the roots as $r=12$ and $r=6$. The general solution is therefore
$$
a_n=Acdot 12^n+B(-6)^n
$$
and you can do the rest.
answered Jan 31 at 21:12
Mike EarnestMike Earnest
27.3k22152
$endgroup$
$begingroup$
Thanks for checking! :)
$endgroup$
– Wesley Strik
Feb 1 at 13:08
add a comment |
$begingroup$
$$a_n=6a_{n-1}+6cdot 12a_{n-2}$$ is correct. From this you get the characteristic polynomial
$$
r^2-6r-6cdot 12
$$
which factors as
$$
(r-12)(r+6)
$$
Note that the roots of this are $r=12$ and $r=-6$. This is where you went wrong; you had the roots as $r=12$ and $r=6$. The general solution is therefore
$$
a_n=Acdot 12^n+B(-6)^n
$$
and you can do the rest.
answered Jan 31 at 21:12
Mike EarnestMike Earnest
27.3k22152
$endgroup$
$begingroup$
Thanks for checking! :)
$endgroup$
– Wesley Strik
Feb 1 at 13:08
add a comment |
$begingroup$
$$a_n=6a_{n-1}+6cdot 12a_{n-2}$$ is correct. From this you get the characteristic polynomial
$$
r^2-6r-6cdot 12
$$
which factors as
$$
(r-12)(r+6)
$$
Note that the roots of this are $r=12$ and $r=-6$. This is where you went wrong; you had the roots as $r=12$ and $r=6$. The general solution is therefore
$$
a_n=Acdot 12^n+B(-6)^n
$$
and you can do the rest.
answered Jan 31 at 21:12
Mike EarnestMike Earnest
27.3k22152
$endgroup$
$$a_n=6a_{n-1}+6cdot 12a_{n-2}$$ is correct. From this you get the characteristic polynomial
$$
r^2-6r-6cdot 12
$$
which factors as
$$
(r-12)(r+6)
$$
Note that the roots of this are $r=12$ and $r=-6$. This is where you went wrong; you had the roots as $r=12$ and $r=6$. The general solution is therefore
$$
a_n=Acdot 12^n+B(-6)^n
$$
and you can do the rest.
answered Jan 31 at 21:12
Mike EarnestMike Earnest
27.3k22152
answered Jan 31 at 21:12
Mike EarnestMike Earnest
27.3k22152
answered Jan 31 at 21:12
Mike EarnestMike Earnest
27.3k22152
answered Jan 31 at 21:12
Mike EarnestMike Earnest
27.3k22152
27.3k22152
$begingroup$
Thanks for checking! :)
$endgroup$
– Wesley Strik
Feb 1 at 13:08
add a comment |
$begingroup$
Thanks for checking! :)
$endgroup$
– Wesley Strik
Feb 1 at 13:08
$begingroup$
Thanks for checking! :)
$endgroup$
– Wesley Strik
Feb 1 at 13:08
$begingroup$
Thanks for checking! :)
$endgroup$
– Wesley Strik
Feb 1 at 13:08
add a comment |
$begingroup$
if there are $a_n$ n-letter words, $c_n$ end in a consonant and $v_n$ end in a vowel
$a_n = c_n + v_n$
To make a n+1 letter word, we take the a_n letter words and tack a letter onto the end.
$a_n = 6c_n + 12 v_n\
c_{n+1} = 12v_n\
v_{n+1} = 6 c_n+ 6v_n = 6 a_n$
We probably could do away with $a_n$ and represent just with $c_n, v_n$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = begin{bmatrix}0&12\6 & 6end{bmatrix}begin{bmatrix} c_{n}\v_{n} end {bmatrix}$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = begin{bmatrix}0&12\6 & 6end{bmatrix}^nbegin{bmatrix} c_{1}\v_{1} end {bmatrix}$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = frac 13 begin{bmatrix}1&-2\1 & 1end{bmatrix}begin{bmatrix}12^n&0\ 0& -6^nend{bmatrix}begin{bmatrix}1&2\-1 & 1end{bmatrix}begin{bmatrix} 12\6 end {bmatrix}$
$a_n = frac {4(12)^n - (-6)^n}{3}$
answered Jan 31 at 21:51
Doug MDoug M
45.4k31954
$endgroup$
$begingroup$
This is a nice method that I will try to use myself more often :)
$endgroup$
– Wesley Strik
Feb 1 at 13:09
add a comment |
$begingroup$
if there are $a_n$ n-letter words, $c_n$ end in a consonant and $v_n$ end in a vowel
$a_n = c_n + v_n$
To make a n+1 letter word, we take the a_n letter words and tack a letter onto the end.
$a_n = 6c_n + 12 v_n\
c_{n+1} = 12v_n\
v_{n+1} = 6 c_n+ 6v_n = 6 a_n$
We probably could do away with $a_n$ and represent just with $c_n, v_n$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = begin{bmatrix}0&12\6 & 6end{bmatrix}begin{bmatrix} c_{n}\v_{n} end {bmatrix}$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = begin{bmatrix}0&12\6 & 6end{bmatrix}^nbegin{bmatrix} c_{1}\v_{1} end {bmatrix}$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = frac 13 begin{bmatrix}1&-2\1 & 1end{bmatrix}begin{bmatrix}12^n&0\ 0& -6^nend{bmatrix}begin{bmatrix}1&2\-1 & 1end{bmatrix}begin{bmatrix} 12\6 end {bmatrix}$
$a_n = frac {4(12)^n - (-6)^n}{3}$
answered Jan 31 at 21:51
Doug MDoug M
45.4k31954
$endgroup$
$begingroup$
This is a nice method that I will try to use myself more often :)
$endgroup$
– Wesley Strik
Feb 1 at 13:09
add a comment |
$begingroup$
if there are $a_n$ n-letter words, $c_n$ end in a consonant and $v_n$ end in a vowel
$a_n = c_n + v_n$
To make a n+1 letter word, we take the a_n letter words and tack a letter onto the end.
$a_n = 6c_n + 12 v_n\
c_{n+1} = 12v_n\
v_{n+1} = 6 c_n+ 6v_n = 6 a_n$
We probably could do away with $a_n$ and represent just with $c_n, v_n$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = begin{bmatrix}0&12\6 & 6end{bmatrix}begin{bmatrix} c_{n}\v_{n} end {bmatrix}$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = begin{bmatrix}0&12\6 & 6end{bmatrix}^nbegin{bmatrix} c_{1}\v_{1} end {bmatrix}$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = frac 13 begin{bmatrix}1&-2\1 & 1end{bmatrix}begin{bmatrix}12^n&0\ 0& -6^nend{bmatrix}begin{bmatrix}1&2\-1 & 1end{bmatrix}begin{bmatrix} 12\6 end {bmatrix}$
$a_n = frac {4(12)^n - (-6)^n}{3}$
answered Jan 31 at 21:51
Doug MDoug M
45.4k31954
$endgroup$
if there are $a_n$ n-letter words, $c_n$ end in a consonant and $v_n$ end in a vowel
$a_n = c_n + v_n$
To make a n+1 letter word, we take the a_n letter words and tack a letter onto the end.
$a_n = 6c_n + 12 v_n\
c_{n+1} = 12v_n\
v_{n+1} = 6 c_n+ 6v_n = 6 a_n$
We probably could do away with $a_n$ and represent just with $c_n, v_n$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = begin{bmatrix}0&12\6 & 6end{bmatrix}begin{bmatrix} c_{n}\v_{n} end {bmatrix}$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = begin{bmatrix}0&12\6 & 6end{bmatrix}^nbegin{bmatrix} c_{1}\v_{1} end {bmatrix}$
$begin{bmatrix} c_{n+1}\v_{n+1} end {bmatrix} = frac 13 begin{bmatrix}1&-2\1 & 1end{bmatrix}begin{bmatrix}12^n&0\ 0& -6^nend{bmatrix}begin{bmatrix}1&2\-1 & 1end{bmatrix}begin{bmatrix} 12\6 end {bmatrix}$
$a_n = frac {4(12)^n - (-6)^n}{3}$
answered Jan 31 at 21:51
Doug MDoug M
45.4k31954
answered Jan 31 at 21:51
Doug MDoug M
45.4k31954
answered Jan 31 at 21:51
Doug MDoug M
45.4k31954
answered Jan 31 at 21:51
Doug MDoug M
45.4k31954
45.4k31954
$begingroup$
This is a nice method that I will try to use myself more often :)
$endgroup$
– Wesley Strik
Feb 1 at 13:09
add a comment |
$begingroup$
This is a nice method that I will try to use myself more often :)
$endgroup$
– Wesley Strik
Feb 1 at 13:09
$begingroup$
This is a nice method that I will try to use myself more often :)
$endgroup$
– Wesley Strik
Feb 1 at 13:09
$begingroup$
This is a nice method that I will try to use myself more often :)
$endgroup$
– Wesley Strik
Feb 1 at 13:09
add a comment |
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$begingroup$
Your reasoning looks fine to me :-)
$endgroup$
– Nicolas FRANCOIS
Jan 31 at 20:54
$begingroup$
You have a typo; you wrote "$r^2=6r-6cdot 16=0$", when I think you meant "$r^2-6r-6cdot 12=0$".
$endgroup$
– Mike Earnest
Jan 31 at 21:33
1
$begingroup$
$a_3=180 cdot 6 + 6 cdot 12 cdot 18 =1080+1296=2376$
$endgroup$
– Daniel Mathias
Feb 1 at 1:37