In the expansion of $ax^3 (2+ax)^{11}$, the coefficient of the term in $x^5$ is $11880$. Find the value of...












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I have been trying this question out for a while now, and somehow still haven't quite got the hang of it.
It is a reverse binomial expansion question, but I am missing out on a few core steps in my working out.



Any help is greatly appreciated :)






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    0














    I have been trying this question out for a while now, and somehow still haven't quite got the hang of it.
    It is a reverse binomial expansion question, but I am missing out on a few core steps in my working out.



    Any help is greatly appreciated :)






    binomial-theorem







    share|cite|improve this question



























      0












      0








      0







      I have been trying this question out for a while now, and somehow still haven't quite got the hang of it.
      It is a reverse binomial expansion question, but I am missing out on a few core steps in my working out.



      Any help is greatly appreciated :)






      binomial-theorem







      share|cite|improve this question















      I have been trying this question out for a while now, and somehow still haven't quite got the hang of it.
      It is a reverse binomial expansion question, but I am missing out on a few core steps in my working out.



      Any help is greatly appreciated :)






      binomial-theorem




      binomial-theorem






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      edited Jan 3 at 19:15









      Luca Bressan

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      asked Jan 3 at 19:10









      mathsllama

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          The term with $x^5$ is $28160x^5a^3$, so we have $440(64a^3-27)=0$, which gives
          $$
          a=frac{3}{4}.
          $$






          share|cite|improve this answer





























            0














            The coefficient of the term in $x^{5}$ is $55*2^{9}a^{3}$ which equals to 11880 then $a={3over 4}$






            share|cite|improve this answer





























              0














              Coefficient of $x^2$ in $(2+ax)^{11}$ is $$binom{11}9cdot2^9cdot a^2=28160a^2$$ so coefficient of $x^5$ in $ax^3 (2+ax)^{11}$ is $$28160a^3=11880implies a^3=frac{27}{64}=left(frac34right)^3impliesboxed{a=frac34}$$






              share|cite|improve this answer





























                0














                Using the binomial theorem:
                $$(2+ax)^{11}=sum_{k=0}^{11}{11choose k}2^{11-k}a^{k}x^{k}$$
                So:
                $$ax^3(2+ax)^11=sum_{k=0}^{11}{11choose k}2^{11-k}a^{k+1}x^{k+3}$$
                Taking $k=2$ yields the $x^5$-term, namely:
                $$55cdot 2^9cdot a^3=11880$$
                So:
                $$a=frac34$$






                share|cite|improve this answer





























                  0














                  Using the binomial theorem, in the expansion of $(p+q)^n$, $(r+1)^{th}$ term $T_{r+1}={^n}C_rp^rq^{n-r}$. So,



                  $$
                  ax^3.({^{11}}C_r{2}^{r}{(ax)}^{11-r})={^{11}}C_r2^r.a^{12-r}.x^{14-r}=11880x^5\implies14-r=5implies r=9
                  $$



                  $$
                  {^{11}}C_9.2^9.a^{3}=11880implies frac{11!}{9!.2!}.2^9.a^3=11880implies a^3=frac{11880}{2^8.11*10}=frac{108*2}{2^9}=6/8=3/4
                  $$






                  share|cite|improve this answer





















                  • Many thanks! :)
                    – mathsllama
                    Jan 3 at 19:41










                  • @mathsllama u could accept appropriate answer.
                    – ss1729
                    Jan 3 at 21:54











                  Your Answer





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                  5 Answers
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                  5 Answers
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                  0














                  The term with $x^5$ is $28160x^5a^3$, so we have $440(64a^3-27)=0$, which gives
                  $$
                  a=frac{3}{4}.
                  $$






                  share|cite|improve this answer


























                    0














                    The term with $x^5$ is $28160x^5a^3$, so we have $440(64a^3-27)=0$, which gives
                    $$
                    a=frac{3}{4}.
                    $$






                    share|cite|improve this answer
























                      0












                      0








                      0






                      The term with $x^5$ is $28160x^5a^3$, so we have $440(64a^3-27)=0$, which gives
                      $$
                      a=frac{3}{4}.
                      $$






                      share|cite|improve this answer












                      The term with $x^5$ is $28160x^5a^3$, so we have $440(64a^3-27)=0$, which gives
                      $$
                      a=frac{3}{4}.
                      $$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 3 at 19:15









                      Dietrich Burde

                      77.4k64386




                      77.4k64386























                          0














                          The coefficient of the term in $x^{5}$ is $55*2^{9}a^{3}$ which equals to 11880 then $a={3over 4}$






                          share|cite|improve this answer


























                            0














                            The coefficient of the term in $x^{5}$ is $55*2^{9}a^{3}$ which equals to 11880 then $a={3over 4}$






                            share|cite|improve this answer
























                              0












                              0








                              0






                              The coefficient of the term in $x^{5}$ is $55*2^{9}a^{3}$ which equals to 11880 then $a={3over 4}$






                              share|cite|improve this answer












                              The coefficient of the term in $x^{5}$ is $55*2^{9}a^{3}$ which equals to 11880 then $a={3over 4}$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 3 at 19:16









                              Mostafa Ayaz

                              13.7k3836




                              13.7k3836























                                  0














                                  Coefficient of $x^2$ in $(2+ax)^{11}$ is $$binom{11}9cdot2^9cdot a^2=28160a^2$$ so coefficient of $x^5$ in $ax^3 (2+ax)^{11}$ is $$28160a^3=11880implies a^3=frac{27}{64}=left(frac34right)^3impliesboxed{a=frac34}$$






                                  share|cite|improve this answer


























                                    0














                                    Coefficient of $x^2$ in $(2+ax)^{11}$ is $$binom{11}9cdot2^9cdot a^2=28160a^2$$ so coefficient of $x^5$ in $ax^3 (2+ax)^{11}$ is $$28160a^3=11880implies a^3=frac{27}{64}=left(frac34right)^3impliesboxed{a=frac34}$$






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      Coefficient of $x^2$ in $(2+ax)^{11}$ is $$binom{11}9cdot2^9cdot a^2=28160a^2$$ so coefficient of $x^5$ in $ax^3 (2+ax)^{11}$ is $$28160a^3=11880implies a^3=frac{27}{64}=left(frac34right)^3impliesboxed{a=frac34}$$






                                      share|cite|improve this answer












                                      Coefficient of $x^2$ in $(2+ax)^{11}$ is $$binom{11}9cdot2^9cdot a^2=28160a^2$$ so coefficient of $x^5$ in $ax^3 (2+ax)^{11}$ is $$28160a^3=11880implies a^3=frac{27}{64}=left(frac34right)^3impliesboxed{a=frac34}$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 3 at 19:16









                                      TheSimpliFire

                                      12k62257




                                      12k62257























                                          0














                                          Using the binomial theorem:
                                          $$(2+ax)^{11}=sum_{k=0}^{11}{11choose k}2^{11-k}a^{k}x^{k}$$
                                          So:
                                          $$ax^3(2+ax)^11=sum_{k=0}^{11}{11choose k}2^{11-k}a^{k+1}x^{k+3}$$
                                          Taking $k=2$ yields the $x^5$-term, namely:
                                          $$55cdot 2^9cdot a^3=11880$$
                                          So:
                                          $$a=frac34$$






                                          share|cite|improve this answer


























                                            0














                                            Using the binomial theorem:
                                            $$(2+ax)^{11}=sum_{k=0}^{11}{11choose k}2^{11-k}a^{k}x^{k}$$
                                            So:
                                            $$ax^3(2+ax)^11=sum_{k=0}^{11}{11choose k}2^{11-k}a^{k+1}x^{k+3}$$
                                            Taking $k=2$ yields the $x^5$-term, namely:
                                            $$55cdot 2^9cdot a^3=11880$$
                                            So:
                                            $$a=frac34$$






                                            share|cite|improve this answer
























                                              0












                                              0








                                              0






                                              Using the binomial theorem:
                                              $$(2+ax)^{11}=sum_{k=0}^{11}{11choose k}2^{11-k}a^{k}x^{k}$$
                                              So:
                                              $$ax^3(2+ax)^11=sum_{k=0}^{11}{11choose k}2^{11-k}a^{k+1}x^{k+3}$$
                                              Taking $k=2$ yields the $x^5$-term, namely:
                                              $$55cdot 2^9cdot a^3=11880$$
                                              So:
                                              $$a=frac34$$






                                              share|cite|improve this answer












                                              Using the binomial theorem:
                                              $$(2+ax)^{11}=sum_{k=0}^{11}{11choose k}2^{11-k}a^{k}x^{k}$$
                                              So:
                                              $$ax^3(2+ax)^11=sum_{k=0}^{11}{11choose k}2^{11-k}a^{k+1}x^{k+3}$$
                                              Taking $k=2$ yields the $x^5$-term, namely:
                                              $$55cdot 2^9cdot a^3=11880$$
                                              So:
                                              $$a=frac34$$







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Jan 3 at 19:20









                                              Mastrem

                                              3,77711230




                                              3,77711230























                                                  0














                                                  Using the binomial theorem, in the expansion of $(p+q)^n$, $(r+1)^{th}$ term $T_{r+1}={^n}C_rp^rq^{n-r}$. So,



                                                  $$
                                                  ax^3.({^{11}}C_r{2}^{r}{(ax)}^{11-r})={^{11}}C_r2^r.a^{12-r}.x^{14-r}=11880x^5\implies14-r=5implies r=9
                                                  $$



                                                  $$
                                                  {^{11}}C_9.2^9.a^{3}=11880implies frac{11!}{9!.2!}.2^9.a^3=11880implies a^3=frac{11880}{2^8.11*10}=frac{108*2}{2^9}=6/8=3/4
                                                  $$






                                                  share|cite|improve this answer





















                                                  • Many thanks! :)
                                                    – mathsllama
                                                    Jan 3 at 19:41










                                                  • @mathsllama u could accept appropriate answer.
                                                    – ss1729
                                                    Jan 3 at 21:54
















                                                  0














                                                  Using the binomial theorem, in the expansion of $(p+q)^n$, $(r+1)^{th}$ term $T_{r+1}={^n}C_rp^rq^{n-r}$. So,



                                                  $$
                                                  ax^3.({^{11}}C_r{2}^{r}{(ax)}^{11-r})={^{11}}C_r2^r.a^{12-r}.x^{14-r}=11880x^5\implies14-r=5implies r=9
                                                  $$



                                                  $$
                                                  {^{11}}C_9.2^9.a^{3}=11880implies frac{11!}{9!.2!}.2^9.a^3=11880implies a^3=frac{11880}{2^8.11*10}=frac{108*2}{2^9}=6/8=3/4
                                                  $$






                                                  share|cite|improve this answer





















                                                  • Many thanks! :)
                                                    – mathsllama
                                                    Jan 3 at 19:41










                                                  • @mathsllama u could accept appropriate answer.
                                                    – ss1729
                                                    Jan 3 at 21:54














                                                  0












                                                  0








                                                  0






                                                  Using the binomial theorem, in the expansion of $(p+q)^n$, $(r+1)^{th}$ term $T_{r+1}={^n}C_rp^rq^{n-r}$. So,



                                                  $$
                                                  ax^3.({^{11}}C_r{2}^{r}{(ax)}^{11-r})={^{11}}C_r2^r.a^{12-r}.x^{14-r}=11880x^5\implies14-r=5implies r=9
                                                  $$



                                                  $$
                                                  {^{11}}C_9.2^9.a^{3}=11880implies frac{11!}{9!.2!}.2^9.a^3=11880implies a^3=frac{11880}{2^8.11*10}=frac{108*2}{2^9}=6/8=3/4
                                                  $$






                                                  share|cite|improve this answer












                                                  Using the binomial theorem, in the expansion of $(p+q)^n$, $(r+1)^{th}$ term $T_{r+1}={^n}C_rp^rq^{n-r}$. So,



                                                  $$
                                                  ax^3.({^{11}}C_r{2}^{r}{(ax)}^{11-r})={^{11}}C_r2^r.a^{12-r}.x^{14-r}=11880x^5\implies14-r=5implies r=9
                                                  $$



                                                  $$
                                                  {^{11}}C_9.2^9.a^{3}=11880implies frac{11!}{9!.2!}.2^9.a^3=11880implies a^3=frac{11880}{2^8.11*10}=frac{108*2}{2^9}=6/8=3/4
                                                  $$







                                                  share|cite|improve this answer












                                                  share|cite|improve this answer



                                                  share|cite|improve this answer










                                                  answered Jan 3 at 19:29









                                                  ss1729

                                                  1,786723




                                                  1,786723












                                                  • Many thanks! :)
                                                    – mathsllama
                                                    Jan 3 at 19:41










                                                  • @mathsllama u could accept appropriate answer.
                                                    – ss1729
                                                    Jan 3 at 21:54


















                                                  • Many thanks! :)
                                                    – mathsllama
                                                    Jan 3 at 19:41










                                                  • @mathsllama u could accept appropriate answer.
                                                    – ss1729
                                                    Jan 3 at 21:54
















                                                  Many thanks! :)
                                                  – mathsllama
                                                  Jan 3 at 19:41




                                                  Many thanks! :)
                                                  – mathsllama
                                                  Jan 3 at 19:41












                                                  @mathsllama u could accept appropriate answer.
                                                  – ss1729
                                                  Jan 3 at 21:54




                                                  @mathsllama u could accept appropriate answer.
                                                  – ss1729
                                                  Jan 3 at 21:54


















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