Binomial Theorem with Three Terms
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$(x^2 + 2 + frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
combinatorics binomial-coefficients
$endgroup$
add a comment |
$begingroup$
$(x^2 + 2 + frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
combinatorics binomial-coefficients
$endgroup$
1
$begingroup$
Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
$endgroup$
– Henry
Nov 19 '18 at 12:16
1
$begingroup$
A "binomial with three terms" should probably be called a trinomial.
$endgroup$
– Torsten Schoeneberg
Nov 20 '18 at 8:42
add a comment |
$begingroup$
$(x^2 + 2 + frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
combinatorics binomial-coefficients
$endgroup$
$(x^2 + 2 + frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
combinatorics binomial-coefficients
combinatorics binomial-coefficients
asked Nov 19 '18 at 10:18
ultralightultralight
786
786
1
$begingroup$
Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
$endgroup$
– Henry
Nov 19 '18 at 12:16
1
$begingroup$
A "binomial with three terms" should probably be called a trinomial.
$endgroup$
– Torsten Schoeneberg
Nov 20 '18 at 8:42
add a comment |
1
$begingroup$
Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
$endgroup$
– Henry
Nov 19 '18 at 12:16
1
$begingroup$
A "binomial with three terms" should probably be called a trinomial.
$endgroup$
– Torsten Schoeneberg
Nov 20 '18 at 8:42
1
1
$begingroup$
Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
$endgroup$
– Henry
Nov 19 '18 at 12:16
$begingroup$
Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
$endgroup$
– Henry
Nov 19 '18 at 12:16
1
1
$begingroup$
A "binomial with three terms" should probably be called a trinomial.
$endgroup$
– Torsten Schoeneberg
Nov 20 '18 at 8:42
$begingroup$
A "binomial with three terms" should probably be called a trinomial.
$endgroup$
– Torsten Schoeneberg
Nov 20 '18 at 8:42
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$
There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.
Thus the coefficient of $x^8$ is $8(35)+21 = 301$
$endgroup$
$begingroup$
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
$endgroup$
– nafhgood
Nov 19 '18 at 10:47
$begingroup$
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
$endgroup$
– Mohammad Riazi-Kermani
Nov 19 '18 at 10:51
add a comment |
$begingroup$
The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
$$
where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.
Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,qquad i+kle 7
$$
Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases
$i=4$, $k=0$, $j=3$;
$i=5$, $k=2$, $j=0$.
Thus the coefficient is
$$
2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
$$
$endgroup$
add a comment |
$begingroup$
Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.
Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.
If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$
If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$
If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$
so the answer is $301$.
$endgroup$
add a comment |
$begingroup$
The answer is 301.
Just trust your plan of the twofold use of the binomial formula:
First step
$$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$
Second step
$$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$
Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$
$$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$
Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.
$endgroup$
add a comment |
$begingroup$
To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.
This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have
7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7
7+4+4, 4+7+4, 4+4+7
6+6+3, 6+3+6, 3+6+6
6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6
That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).
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What about to undelete that OP?
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– gimusi
Nov 26 '18 at 9:33
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$
There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.
Thus the coefficient of $x^8$ is $8(35)+21 = 301$
$endgroup$
$begingroup$
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
$endgroup$
– nafhgood
Nov 19 '18 at 10:47
$begingroup$
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
$endgroup$
– Mohammad Riazi-Kermani
Nov 19 '18 at 10:51
add a comment |
$begingroup$
In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$
There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.
Thus the coefficient of $x^8$ is $8(35)+21 = 301$
$endgroup$
$begingroup$
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
$endgroup$
– nafhgood
Nov 19 '18 at 10:47
$begingroup$
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
$endgroup$
– Mohammad Riazi-Kermani
Nov 19 '18 at 10:51
add a comment |
$begingroup$
In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$
There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.
Thus the coefficient of $x^8$ is $8(35)+21 = 301$
$endgroup$
In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$
There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.
Thus the coefficient of $x^8$ is $8(35)+21 = 301$
answered Nov 19 '18 at 10:44
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
$begingroup$
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
$endgroup$
– nafhgood
Nov 19 '18 at 10:47
$begingroup$
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
$endgroup$
– Mohammad Riazi-Kermani
Nov 19 '18 at 10:51
add a comment |
$begingroup$
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
$endgroup$
– nafhgood
Nov 19 '18 at 10:47
$begingroup$
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
$endgroup$
– Mohammad Riazi-Kermani
Nov 19 '18 at 10:51
$begingroup$
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
$endgroup$
– nafhgood
Nov 19 '18 at 10:47
$begingroup$
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
$endgroup$
– nafhgood
Nov 19 '18 at 10:47
$begingroup$
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
$endgroup$
– Mohammad Riazi-Kermani
Nov 19 '18 at 10:51
$begingroup$
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
$endgroup$
– Mohammad Riazi-Kermani
Nov 19 '18 at 10:51
add a comment |
$begingroup$
The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
$$
where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.
Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,qquad i+kle 7
$$
Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases
$i=4$, $k=0$, $j=3$;
$i=5$, $k=2$, $j=0$.
Thus the coefficient is
$$
2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
$$
$endgroup$
add a comment |
$begingroup$
The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
$$
where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.
Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,qquad i+kle 7
$$
Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases
$i=4$, $k=0$, $j=3$;
$i=5$, $k=2$, $j=0$.
Thus the coefficient is
$$
2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
$$
$endgroup$
add a comment |
$begingroup$
The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
$$
where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.
Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,qquad i+kle 7
$$
Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases
$i=4$, $k=0$, $j=3$;
$i=5$, $k=2$, $j=0$.
Thus the coefficient is
$$
2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
$$
$endgroup$
The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
$$
where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.
Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,qquad i+kle 7
$$
Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases
$i=4$, $k=0$, $j=3$;
$i=5$, $k=2$, $j=0$.
Thus the coefficient is
$$
2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
$$
edited Nov 19 '18 at 21:13
answered Nov 19 '18 at 15:26
egregegreg
182k1485203
182k1485203
add a comment |
add a comment |
$begingroup$
Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.
Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.
If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$
If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$
If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$
so the answer is $301$.
$endgroup$
add a comment |
$begingroup$
Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.
Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.
If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$
If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$
If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$
so the answer is $301$.
$endgroup$
add a comment |
$begingroup$
Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.
Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.
If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$
If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$
If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$
so the answer is $301$.
$endgroup$
Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.
Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.
If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$
If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$
If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$
so the answer is $301$.
edited Nov 19 '18 at 15:37
Rad80
30718
30718
answered Nov 19 '18 at 10:42
greedoidgreedoid
42.4k1153105
42.4k1153105
add a comment |
add a comment |
$begingroup$
The answer is 301.
Just trust your plan of the twofold use of the binomial formula:
First step
$$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$
Second step
$$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$
Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$
$$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$
Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.
$endgroup$
add a comment |
$begingroup$
The answer is 301.
Just trust your plan of the twofold use of the binomial formula:
First step
$$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$
Second step
$$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$
Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$
$$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$
Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.
$endgroup$
add a comment |
$begingroup$
The answer is 301.
Just trust your plan of the twofold use of the binomial formula:
First step
$$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$
Second step
$$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$
Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$
$$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$
Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.
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The answer is 301.
Just trust your plan of the twofold use of the binomial formula:
First step
$$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$
Second step
$$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$
Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$
$$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$
Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.
edited Nov 19 '18 at 11:29
answered Nov 19 '18 at 11:09
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,425618
3,425618
add a comment |
add a comment |
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To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.
This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have
7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7
7+4+4, 4+7+4, 4+4+7
6+6+3, 6+3+6, 3+6+6
6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6
That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).
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What about to undelete that OP?
$endgroup$
– gimusi
Nov 26 '18 at 9:33
add a comment |
$begingroup$
To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.
This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have
7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7
7+4+4, 4+7+4, 4+4+7
6+6+3, 6+3+6, 3+6+6
6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6
That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).
$endgroup$
$begingroup$
What about to undelete that OP?
$endgroup$
– gimusi
Nov 26 '18 at 9:33
add a comment |
$begingroup$
To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.
This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have
7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7
7+4+4, 4+7+4, 4+4+7
6+6+3, 6+3+6, 3+6+6
6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6
That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).
$endgroup$
To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.
This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have
7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7
7+4+4, 4+7+4, 4+4+7
6+6+3, 6+3+6, 3+6+6
6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6
That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).
answered Nov 19 '18 at 18:37
AcccumulationAcccumulation
6,9642618
6,9642618
$begingroup$
What about to undelete that OP?
$endgroup$
– gimusi
Nov 26 '18 at 9:33
add a comment |
$begingroup$
What about to undelete that OP?
$endgroup$
– gimusi
Nov 26 '18 at 9:33
$begingroup$
What about to undelete that OP?
$endgroup$
– gimusi
Nov 26 '18 at 9:33
$begingroup$
What about to undelete that OP?
$endgroup$
– gimusi
Nov 26 '18 at 9:33
add a comment |
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Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
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– Henry
Nov 19 '18 at 12:16
1
$begingroup$
A "binomial with three terms" should probably be called a trinomial.
$endgroup$
– Torsten Schoeneberg
Nov 20 '18 at 8:42