What is the coefficient of $x^{11}$ in $(3x-9)^{19}$?












1












$begingroup$


I am currently studying for finals, and I do not know how to do this problem from my study guide. I have tried to watch a few YouTube videos and I know that I will end up with $3x^{11} times (-9)^8$, but from here I do not know where to go. The YouTube videos I have watched have been unclear, and I do not have a chance to ask my professor until just a little before the exam. If anyone has any tips on where to go from here I would greatly appreciate it!






exponential-function binomial-coefficients binomial-theorem







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$endgroup$












  • $begingroup$
    Don't forget about the binomial coefficient.
    $endgroup$
    – John Wayland Bales
    Dec 7 '18 at 16:40










  • $begingroup$
    Are you aware of the BInomial expansion ${(ax-b)}^n = sum ^ n _ {r=0} binom{n}{r} {(ax)}^{n-r} cdot (-1)^r cdot b^r$?
    $endgroup$
    – Prakhar Nagpal
    Dec 7 '18 at 16:41


















1












$begingroup$


I am currently studying for finals, and I do not know how to do this problem from my study guide. I have tried to watch a few YouTube videos and I know that I will end up with $3x^{11} times (-9)^8$, but from here I do not know where to go. The YouTube videos I have watched have been unclear, and I do not have a chance to ask my professor until just a little before the exam. If anyone has any tips on where to go from here I would greatly appreciate it!






exponential-function binomial-coefficients binomial-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    Don't forget about the binomial coefficient.
    $endgroup$
    – John Wayland Bales
    Dec 7 '18 at 16:40










  • $begingroup$
    Are you aware of the BInomial expansion ${(ax-b)}^n = sum ^ n _ {r=0} binom{n}{r} {(ax)}^{n-r} cdot (-1)^r cdot b^r$?
    $endgroup$
    – Prakhar Nagpal
    Dec 7 '18 at 16:41
















1












1








1





$begingroup$


I am currently studying for finals, and I do not know how to do this problem from my study guide. I have tried to watch a few YouTube videos and I know that I will end up with $3x^{11} times (-9)^8$, but from here I do not know where to go. The YouTube videos I have watched have been unclear, and I do not have a chance to ask my professor until just a little before the exam. If anyone has any tips on where to go from here I would greatly appreciate it!






exponential-function binomial-coefficients binomial-theorem







share|cite|improve this question











$endgroup$




I am currently studying for finals, and I do not know how to do this problem from my study guide. I have tried to watch a few YouTube videos and I know that I will end up with $3x^{11} times (-9)^8$, but from here I do not know where to go. The YouTube videos I have watched have been unclear, and I do not have a chance to ask my professor until just a little before the exam. If anyone has any tips on where to go from here I would greatly appreciate it!






exponential-function binomial-coefficients binomial-theorem




exponential-function binomial-coefficients binomial-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 17:10









PM.

3,3532925




3,3532925










asked Dec 7 '18 at 16:36









bmurf17bmurf17

104




104












  • $begingroup$
    Don't forget about the binomial coefficient.
    $endgroup$
    – John Wayland Bales
    Dec 7 '18 at 16:40










  • $begingroup$
    Are you aware of the BInomial expansion ${(ax-b)}^n = sum ^ n _ {r=0} binom{n}{r} {(ax)}^{n-r} cdot (-1)^r cdot b^r$?
    $endgroup$
    – Prakhar Nagpal
    Dec 7 '18 at 16:41




















  • $begingroup$
    Don't forget about the binomial coefficient.
    $endgroup$
    – John Wayland Bales
    Dec 7 '18 at 16:40










  • $begingroup$
    Are you aware of the BInomial expansion ${(ax-b)}^n = sum ^ n _ {r=0} binom{n}{r} {(ax)}^{n-r} cdot (-1)^r cdot b^r$?
    $endgroup$
    – Prakhar Nagpal
    Dec 7 '18 at 16:41


















$begingroup$
Don't forget about the binomial coefficient.
$endgroup$
– John Wayland Bales
Dec 7 '18 at 16:40




$begingroup$
Don't forget about the binomial coefficient.
$endgroup$
– John Wayland Bales
Dec 7 '18 at 16:40












$begingroup$
Are you aware of the BInomial expansion ${(ax-b)}^n = sum ^ n _ {r=0} binom{n}{r} {(ax)}^{n-r} cdot (-1)^r cdot b^r$?
$endgroup$
– Prakhar Nagpal
Dec 7 '18 at 16:41






$begingroup$
Are you aware of the BInomial expansion ${(ax-b)}^n = sum ^ n _ {r=0} binom{n}{r} {(ax)}^{n-r} cdot (-1)^r cdot b^r$?
$endgroup$
– Prakhar Nagpal
Dec 7 '18 at 16:41












4 Answers
4






active

oldest

votes


















1












$begingroup$

$$(3x-9)^{19} = (3x-9)(3x-9) cdots (3x-9) $$



Notice that on the right hand side there exist $19$ products. A term can be formed by choosing either $3x$ or $-9$ from each of these $19$ products.



If your goal is to find the coefficient of $x^{11}$, you have to choose $3x$ for a $11$ times from the available $19$ products, and then choose $-9$ for $8$ times from the remaining $8$ products.



If you choose $3x$ for $11$ times, you end up with $color{blue}{(3x)^{11}}$, and you can do this in exactly $color{green}{binom{19}{11}}$ ways. $color{grey}{text{(because you have $19$ products available and you're considering only $11$ of them for $3x$.)}}$



For each of the above $color{green}{binom{19}{11}}$ ways, you can choose $-9$ from the remaining $8$ products, this gives $color{red}{(-9)^8}$

So the overall term would be $color{green}{binom{19}{11}}color{blue}{(3x)^{11}}color{red}{(-9)^8}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you this makes sense! I completely forgot about the choose
    $endgroup$
    – bmurf17
    Dec 7 '18 at 19:04



















0












$begingroup$

HINT:
The Binomial expansion : $$(a+b)^{n}= sum_{k=0}^{n} {n choose k}{a}^{k}{b}^{n-k}.$$



Substitute $a=3x$ and $b=-9$ . To find out the coefficient of $x^{11} $,put $k=11$ in ${n choose k}{a}^{k}{b}^{n-k}$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Note that by expansion of a binomial to the $n^{th}$ power, you get



    $$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r = a^n+a^{n-1}b+a^{n-2}b^2+…+a^{n-r}b^r+…+ab^{n-1}+b^n$$



    where ${n choose r}$ corresponds to the binomial coefficients, or Pascal Numbers. (Never forget those for coefficients.)



    For $(3x-9)^{19}$, the first term has $x^{19}$, the second term has $x^{18}$, etc. So, $x^{11}$ corresponds to the ninth term. Hence, you get



    $${19 choose 19-8} (3x)^{11}(-9)^8$$



    $${19 choose 11} (3x)^{11}9^8$$



    Simplifying from here, you can solve for the ninth term and find the coefficient of $x^{11}$.






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      You are part way there.



      If you were to write out $(3x-9)^{19}$ in full, it would be a product of nineteen $ (3x-9)$. To get an $x^{11}$ you would need to pick 11 of these $(3x-9)$.



      There are $19 choose 11$ ways of doing that, so the term in $x^{11}$ is
      $$
      {19 choose 11} (3x)^{11} (-9)^8
      $$



      Thus the coefficient of $x^{11}$ is



      $$
      {19 choose 11} (3)^{11} (-9)^8 = {19 choose 11}3^{27}
      $$






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        $$(3x-9)^{19} = (3x-9)(3x-9) cdots (3x-9) $$



        Notice that on the right hand side there exist $19$ products. A term can be formed by choosing either $3x$ or $-9$ from each of these $19$ products.



        If your goal is to find the coefficient of $x^{11}$, you have to choose $3x$ for a $11$ times from the available $19$ products, and then choose $-9$ for $8$ times from the remaining $8$ products.



        If you choose $3x$ for $11$ times, you end up with $color{blue}{(3x)^{11}}$, and you can do this in exactly $color{green}{binom{19}{11}}$ ways. $color{grey}{text{(because you have $19$ products available and you're considering only $11$ of them for $3x$.)}}$



        For each of the above $color{green}{binom{19}{11}}$ ways, you can choose $-9$ from the remaining $8$ products, this gives $color{red}{(-9)^8}$

        So the overall term would be $color{green}{binom{19}{11}}color{blue}{(3x)^{11}}color{red}{(-9)^8}$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Thank you this makes sense! I completely forgot about the choose
          $endgroup$
          – bmurf17
          Dec 7 '18 at 19:04
















        1












        $begingroup$

        $$(3x-9)^{19} = (3x-9)(3x-9) cdots (3x-9) $$



        Notice that on the right hand side there exist $19$ products. A term can be formed by choosing either $3x$ or $-9$ from each of these $19$ products.



        If your goal is to find the coefficient of $x^{11}$, you have to choose $3x$ for a $11$ times from the available $19$ products, and then choose $-9$ for $8$ times from the remaining $8$ products.



        If you choose $3x$ for $11$ times, you end up with $color{blue}{(3x)^{11}}$, and you can do this in exactly $color{green}{binom{19}{11}}$ ways. $color{grey}{text{(because you have $19$ products available and you're considering only $11$ of them for $3x$.)}}$



        For each of the above $color{green}{binom{19}{11}}$ ways, you can choose $-9$ from the remaining $8$ products, this gives $color{red}{(-9)^8}$

        So the overall term would be $color{green}{binom{19}{11}}color{blue}{(3x)^{11}}color{red}{(-9)^8}$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Thank you this makes sense! I completely forgot about the choose
          $endgroup$
          – bmurf17
          Dec 7 '18 at 19:04














        1












        1








        1





        $begingroup$

        $$(3x-9)^{19} = (3x-9)(3x-9) cdots (3x-9) $$



        Notice that on the right hand side there exist $19$ products. A term can be formed by choosing either $3x$ or $-9$ from each of these $19$ products.



        If your goal is to find the coefficient of $x^{11}$, you have to choose $3x$ for a $11$ times from the available $19$ products, and then choose $-9$ for $8$ times from the remaining $8$ products.



        If you choose $3x$ for $11$ times, you end up with $color{blue}{(3x)^{11}}$, and you can do this in exactly $color{green}{binom{19}{11}}$ ways. $color{grey}{text{(because you have $19$ products available and you're considering only $11$ of them for $3x$.)}}$



        For each of the above $color{green}{binom{19}{11}}$ ways, you can choose $-9$ from the remaining $8$ products, this gives $color{red}{(-9)^8}$

        So the overall term would be $color{green}{binom{19}{11}}color{blue}{(3x)^{11}}color{red}{(-9)^8}$






        share|cite|improve this answer











        $endgroup$



        $$(3x-9)^{19} = (3x-9)(3x-9) cdots (3x-9) $$



        Notice that on the right hand side there exist $19$ products. A term can be formed by choosing either $3x$ or $-9$ from each of these $19$ products.



        If your goal is to find the coefficient of $x^{11}$, you have to choose $3x$ for a $11$ times from the available $19$ products, and then choose $-9$ for $8$ times from the remaining $8$ products.



        If you choose $3x$ for $11$ times, you end up with $color{blue}{(3x)^{11}}$, and you can do this in exactly $color{green}{binom{19}{11}}$ ways. $color{grey}{text{(because you have $19$ products available and you're considering only $11$ of them for $3x$.)}}$



        For each of the above $color{green}{binom{19}{11}}$ ways, you can choose $-9$ from the remaining $8$ products, this gives $color{red}{(-9)^8}$

        So the overall term would be $color{green}{binom{19}{11}}color{blue}{(3x)^{11}}color{red}{(-9)^8}$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 7 '18 at 17:12

























        answered Dec 7 '18 at 17:00









        rsadhvikarsadhvika

        1,7101228




        1,7101228












        • $begingroup$
          Thank you this makes sense! I completely forgot about the choose
          $endgroup$
          – bmurf17
          Dec 7 '18 at 19:04


















        • $begingroup$
          Thank you this makes sense! I completely forgot about the choose
          $endgroup$
          – bmurf17
          Dec 7 '18 at 19:04
















        $begingroup$
        Thank you this makes sense! I completely forgot about the choose
        $endgroup$
        – bmurf17
        Dec 7 '18 at 19:04




        $begingroup$
        Thank you this makes sense! I completely forgot about the choose
        $endgroup$
        – bmurf17
        Dec 7 '18 at 19:04











        0












        $begingroup$

        HINT:
        The Binomial expansion : $$(a+b)^{n}= sum_{k=0}^{n} {n choose k}{a}^{k}{b}^{n-k}.$$



        Substitute $a=3x$ and $b=-9$ . To find out the coefficient of $x^{11} $,put $k=11$ in ${n choose k}{a}^{k}{b}^{n-k}$






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          HINT:
          The Binomial expansion : $$(a+b)^{n}= sum_{k=0}^{n} {n choose k}{a}^{k}{b}^{n-k}.$$



          Substitute $a=3x$ and $b=-9$ . To find out the coefficient of $x^{11} $,put $k=11$ in ${n choose k}{a}^{k}{b}^{n-k}$






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            HINT:
            The Binomial expansion : $$(a+b)^{n}= sum_{k=0}^{n} {n choose k}{a}^{k}{b}^{n-k}.$$



            Substitute $a=3x$ and $b=-9$ . To find out the coefficient of $x^{11} $,put $k=11$ in ${n choose k}{a}^{k}{b}^{n-k}$






            share|cite|improve this answer









            $endgroup$



            HINT:
            The Binomial expansion : $$(a+b)^{n}= sum_{k=0}^{n} {n choose k}{a}^{k}{b}^{n-k}.$$



            Substitute $a=3x$ and $b=-9$ . To find out the coefficient of $x^{11} $,put $k=11$ in ${n choose k}{a}^{k}{b}^{n-k}$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 7 '18 at 16:48









            Thomas ShelbyThomas Shelby

            3,9542625




            3,9542625























                0












                $begingroup$

                Note that by expansion of a binomial to the $n^{th}$ power, you get



                $$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r = a^n+a^{n-1}b+a^{n-2}b^2+…+a^{n-r}b^r+…+ab^{n-1}+b^n$$



                where ${n choose r}$ corresponds to the binomial coefficients, or Pascal Numbers. (Never forget those for coefficients.)



                For $(3x-9)^{19}$, the first term has $x^{19}$, the second term has $x^{18}$, etc. So, $x^{11}$ corresponds to the ninth term. Hence, you get



                $${19 choose 19-8} (3x)^{11}(-9)^8$$



                $${19 choose 11} (3x)^{11}9^8$$



                Simplifying from here, you can solve for the ninth term and find the coefficient of $x^{11}$.






                share|cite|improve this answer











                $endgroup$


















                  0












                  $begingroup$

                  Note that by expansion of a binomial to the $n^{th}$ power, you get



                  $$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r = a^n+a^{n-1}b+a^{n-2}b^2+…+a^{n-r}b^r+…+ab^{n-1}+b^n$$



                  where ${n choose r}$ corresponds to the binomial coefficients, or Pascal Numbers. (Never forget those for coefficients.)



                  For $(3x-9)^{19}$, the first term has $x^{19}$, the second term has $x^{18}$, etc. So, $x^{11}$ corresponds to the ninth term. Hence, you get



                  $${19 choose 19-8} (3x)^{11}(-9)^8$$



                  $${19 choose 11} (3x)^{11}9^8$$



                  Simplifying from here, you can solve for the ninth term and find the coefficient of $x^{11}$.






                  share|cite|improve this answer











                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Note that by expansion of a binomial to the $n^{th}$ power, you get



                    $$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r = a^n+a^{n-1}b+a^{n-2}b^2+…+a^{n-r}b^r+…+ab^{n-1}+b^n$$



                    where ${n choose r}$ corresponds to the binomial coefficients, or Pascal Numbers. (Never forget those for coefficients.)



                    For $(3x-9)^{19}$, the first term has $x^{19}$, the second term has $x^{18}$, etc. So, $x^{11}$ corresponds to the ninth term. Hence, you get



                    $${19 choose 19-8} (3x)^{11}(-9)^8$$



                    $${19 choose 11} (3x)^{11}9^8$$



                    Simplifying from here, you can solve for the ninth term and find the coefficient of $x^{11}$.






                    share|cite|improve this answer











                    $endgroup$



                    Note that by expansion of a binomial to the $n^{th}$ power, you get



                    $$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r = a^n+a^{n-1}b+a^{n-2}b^2+…+a^{n-r}b^r+…+ab^{n-1}+b^n$$



                    where ${n choose r}$ corresponds to the binomial coefficients, or Pascal Numbers. (Never forget those for coefficients.)



                    For $(3x-9)^{19}$, the first term has $x^{19}$, the second term has $x^{18}$, etc. So, $x^{11}$ corresponds to the ninth term. Hence, you get



                    $${19 choose 19-8} (3x)^{11}(-9)^8$$



                    $${19 choose 11} (3x)^{11}9^8$$



                    Simplifying from here, you can solve for the ninth term and find the coefficient of $x^{11}$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 7 '18 at 17:02

























                    answered Dec 7 '18 at 16:44









                    KM101KM101

                    6,0901525




                    6,0901525























                        0












                        $begingroup$

                        You are part way there.



                        If you were to write out $(3x-9)^{19}$ in full, it would be a product of nineteen $ (3x-9)$. To get an $x^{11}$ you would need to pick 11 of these $(3x-9)$.



                        There are $19 choose 11$ ways of doing that, so the term in $x^{11}$ is
                        $$
                        {19 choose 11} (3x)^{11} (-9)^8
                        $$



                        Thus the coefficient of $x^{11}$ is



                        $$
                        {19 choose 11} (3)^{11} (-9)^8 = {19 choose 11}3^{27}
                        $$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          You are part way there.



                          If you were to write out $(3x-9)^{19}$ in full, it would be a product of nineteen $ (3x-9)$. To get an $x^{11}$ you would need to pick 11 of these $(3x-9)$.



                          There are $19 choose 11$ ways of doing that, so the term in $x^{11}$ is
                          $$
                          {19 choose 11} (3x)^{11} (-9)^8
                          $$



                          Thus the coefficient of $x^{11}$ is



                          $$
                          {19 choose 11} (3)^{11} (-9)^8 = {19 choose 11}3^{27}
                          $$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            You are part way there.



                            If you were to write out $(3x-9)^{19}$ in full, it would be a product of nineteen $ (3x-9)$. To get an $x^{11}$ you would need to pick 11 of these $(3x-9)$.



                            There are $19 choose 11$ ways of doing that, so the term in $x^{11}$ is
                            $$
                            {19 choose 11} (3x)^{11} (-9)^8
                            $$



                            Thus the coefficient of $x^{11}$ is



                            $$
                            {19 choose 11} (3)^{11} (-9)^8 = {19 choose 11}3^{27}
                            $$






                            share|cite|improve this answer









                            $endgroup$



                            You are part way there.



                            If you were to write out $(3x-9)^{19}$ in full, it would be a product of nineteen $ (3x-9)$. To get an $x^{11}$ you would need to pick 11 of these $(3x-9)$.



                            There are $19 choose 11$ ways of doing that, so the term in $x^{11}$ is
                            $$
                            {19 choose 11} (3x)^{11} (-9)^8
                            $$



                            Thus the coefficient of $x^{11}$ is



                            $$
                            {19 choose 11} (3)^{11} (-9)^8 = {19 choose 11}3^{27}
                            $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 7 '18 at 17:04









                            PM.PM.

                            3,3532925




                            3,3532925






























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                                How to change which sound is reproduced for terminal bell?

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                                3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
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                                Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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                                up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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                                Can I use Tabulator js library in my java Spring + Thymeleaf project?

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                                0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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