Find Coefficient of Trinomial Where Term has a Coefficient
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Given a problem such as "find the coefficient of $a^2b^6$ for $(a+3b+2)^10$," how would I go about doing this?
I know the multinomial theorem, but I'm not sure how to approach this problem given that $b$ has a coefficient, as well as the fact that the powers of $a$ and $b$ do not add up to 10.
combinatorics multinomial-coefficients multinomial-theorem
asked Sep 7 at 1:57
hopelessundergrad
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up vote
2
down vote
favorite
Given a problem such as "find the coefficient of $a^2b^6$ for $(a+3b+2)^10$," how would I go about doing this?
I know the multinomial theorem, but I'm not sure how to approach this problem given that $b$ has a coefficient, as well as the fact that the powers of $a$ and $b$ do not add up to 10.
combinatorics multinomial-coefficients multinomial-theorem
asked Sep 7 at 1:57
hopelessundergrad
111
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given a problem such as "find the coefficient of $a^2b^6$ for $(a+3b+2)^10$," how would I go about doing this?
I know the multinomial theorem, but I'm not sure how to approach this problem given that $b$ has a coefficient, as well as the fact that the powers of $a$ and $b$ do not add up to 10.
combinatorics multinomial-coefficients multinomial-theorem
asked Sep 7 at 1:57
hopelessundergrad
111
Given a problem such as "find the coefficient of $a^2b^6$ for $(a+3b+2)^10$," how would I go about doing this?
I know the multinomial theorem, but I'm not sure how to approach this problem given that $b$ has a coefficient, as well as the fact that the powers of $a$ and $b$ do not add up to 10.
combinatorics multinomial-coefficients multinomial-theorem
combinatorics multinomial-coefficients multinomial-theorem
asked Sep 7 at 1:57
hopelessundergrad
111
asked Sep 7 at 1:57
hopelessundergrad
111
asked Sep 7 at 1:57
hopelessundergrad
111
asked Sep 7 at 1:57
hopelessundergrad
111
asked Sep 7 at 1:57
hopelessundergrad
111
111
add a comment |Â
add a comment |Â
1 Answer
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One step at a time, $;big(a+(3b+2)big)^10 = sum_k=0^10 binom10ka^k(3b+2)^10-k,$. The only term containing $,a^2,$ is $,binom102a^2(3b+2)^8,$. Now expand $,(3b+2)^8=sum_k=0^8binom8k(3b)^k2^8-k,$, and note that the only term containing $,b^6,$ is $,binom86(3b)^62^2,$. Next, put the two together.
Or, use the multinomial expansion directly, where the coefficient of $a^2(3b)^62^2$ is $binom10,2, ,6, ,2,,$.
answered Sep 7 at 2:08
dxiv
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
One step at a time, $;big(a+(3b+2)big)^10 = sum_k=0^10 binom10ka^k(3b+2)^10-k,$. The only term containing $,a^2,$ is $,binom102a^2(3b+2)^8,$. Now expand $,(3b+2)^8=sum_k=0^8binom8k(3b)^k2^8-k,$, and note that the only term containing $,b^6,$ is $,binom86(3b)^62^2,$. Next, put the two together.
Or, use the multinomial expansion directly, where the coefficient of $a^2(3b)^62^2$ is $binom10,2, ,6, ,2,,$.
answered Sep 7 at 2:08
dxiv
56.2k64798
add a comment |Â
up vote
1
down vote
One step at a time, $;big(a+(3b+2)big)^10 = sum_k=0^10 binom10ka^k(3b+2)^10-k,$. The only term containing $,a^2,$ is $,binom102a^2(3b+2)^8,$. Now expand $,(3b+2)^8=sum_k=0^8binom8k(3b)^k2^8-k,$, and note that the only term containing $,b^6,$ is $,binom86(3b)^62^2,$. Next, put the two together.
Or, use the multinomial expansion directly, where the coefficient of $a^2(3b)^62^2$ is $binom10,2, ,6, ,2,,$.
answered Sep 7 at 2:08
dxiv
56.2k64798
add a comment |Â
up vote
1
down vote
up vote
1
down vote
One step at a time, $;big(a+(3b+2)big)^10 = sum_k=0^10 binom10ka^k(3b+2)^10-k,$. The only term containing $,a^2,$ is $,binom102a^2(3b+2)^8,$. Now expand $,(3b+2)^8=sum_k=0^8binom8k(3b)^k2^8-k,$, and note that the only term containing $,b^6,$ is $,binom86(3b)^62^2,$. Next, put the two together.
Or, use the multinomial expansion directly, where the coefficient of $a^2(3b)^62^2$ is $binom10,2, ,6, ,2,,$.
answered Sep 7 at 2:08
dxiv
56.2k64798
One step at a time, $;big(a+(3b+2)big)^10 = sum_k=0^10 binom10ka^k(3b+2)^10-k,$. The only term containing $,a^2,$ is $,binom102a^2(3b+2)^8,$. Now expand $,(3b+2)^8=sum_k=0^8binom8k(3b)^k2^8-k,$, and note that the only term containing $,b^6,$ is $,binom86(3b)^62^2,$. Next, put the two together.
Or, use the multinomial expansion directly, where the coefficient of $a^2(3b)^62^2$ is $binom10,2, ,6, ,2,,$.
answered Sep 7 at 2:08
dxiv
56.2k64798
answered Sep 7 at 2:08
dxiv
56.2k64798
answered Sep 7 at 2:08
dxiv
56.2k64798
answered Sep 7 at 2:08
dxiv
56.2k64798
56.2k64798
add a comment |Â
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