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

-1

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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 :)

binomial-theorem

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edited Jan 3 '18 at 19:15

Luca Bressan

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asked Jan 3 '18 at 19:10

mathsllamamathsllama

11

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closed as off-topic by TheSimpliFire, Eevee Trainer, Leucippus, Chris Custer, A. Pongrácz Jan 16 at 7:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Eevee Trainer, Leucippus, Chris Custer, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

-1

$begingroup$

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

edited Jan 3 '18 at 19:15

Luca Bressan

3,92621037

asked Jan 3 '18 at 19:10

mathsllamamathsllama

11

$endgroup$

closed as off-topic by TheSimpliFire, Eevee Trainer, Leucippus, Chris Custer, A. Pongrácz Jan 16 at 7:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Eevee Trainer, Leucippus, Chris Custer, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

-1

-1

-1

$begingroup$

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

edited Jan 3 '18 at 19:15

Luca Bressan

3,92621037

asked Jan 3 '18 at 19:10

mathsllamamathsllama

11

$endgroup$

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

share|cite|improve this question

edited Jan 3 '18 at 19:15

Luca Bressan

3,92621037

asked Jan 3 '18 at 19:10

mathsllamamathsllama

11

share|cite|improve this question

edited Jan 3 '18 at 19:15

Luca Bressan

3,92621037

asked Jan 3 '18 at 19:10

mathsllamamathsllama

11

share|cite|improve this question

share|cite|improve this question

edited Jan 3 '18 at 19:15

Luca Bressan

3,92621037

edited Jan 3 '18 at 19:15

Luca Bressan

3,92621037

edited Jan 3 '18 at 19:15

Luca Bressan

3,92621037

3,92621037

asked Jan 3 '18 at 19:10

mathsllamamathsllama

11

asked Jan 3 '18 at 19:10

mathsllamamathsllama

11

asked Jan 3 '18 at 19:10

mathsllamamathsllama

11

11

closed as off-topic by TheSimpliFire, Eevee Trainer, Leucippus, Chris Custer, A. Pongrácz Jan 16 at 7:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Eevee Trainer, Leucippus, Chris Custer, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by TheSimpliFire, Eevee Trainer, Leucippus, Chris Custer, A. Pongrácz Jan 16 at 7:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Eevee Trainer, Leucippus, Chris Custer, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

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

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answered Jan 3 '18 at 19:15

Dietrich BurdeDietrich Burde

78.4k64386

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0

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

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answered Jan 3 '18 at 19:16

Mostafa AyazMostafa Ayaz

15.3k3939

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0

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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 $$28160a^3=11880implies a^3=frac{27}{64}=left(frac34right)^3impliesboxed{a=frac34}$$

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answered Jan 3 '18 at 19:16

TheSimpliFireTheSimpliFire

12.5k62460

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0

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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:
$$a=frac34$$

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answered Jan 3 '18 at 19:20

MastremMastrem

3,79211230

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add a comment | 

0

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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,

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

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

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answered Jan 3 '18 at 19:29

ss1729ss1729

1,8921723

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Many thanks! :)
  $endgroup$
  – mathsllama
  Jan 3 '18 at 19:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathsllama u could accept appropriate answer.
  $endgroup$
  – ss1729
  Jan 3 '18 at 21:54
  

  
  
  
  

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

0

$begingroup$

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

answered Jan 3 '18 at 19:15

Dietrich BurdeDietrich Burde

78.4k64386

$endgroup$

add a comment | 

0

$begingroup$

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

answered Jan 3 '18 at 19:15

Dietrich BurdeDietrich Burde

78.4k64386

$endgroup$

add a comment | 

0

0

0

$begingroup$

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

answered Jan 3 '18 at 19:15

Dietrich BurdeDietrich Burde

78.4k64386

$endgroup$

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

answered Jan 3 '18 at 19:15

Dietrich BurdeDietrich Burde

78.4k64386

share|cite|improve this answer

share|cite|improve this answer

answered Jan 3 '18 at 19:15

Dietrich BurdeDietrich Burde

78.4k64386

answered Jan 3 '18 at 19:15

Dietrich BurdeDietrich Burde

78.4k64386

answered Jan 3 '18 at 19:15

Dietrich BurdeDietrich Burde

78.4k64386

78.4k64386

add a comment | 

add a comment | 

0

$begingroup$

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

answered Jan 3 '18 at 19:16

Mostafa AyazMostafa Ayaz

15.3k3939

$endgroup$

add a comment | 

0

$begingroup$

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

answered Jan 3 '18 at 19:16

Mostafa AyazMostafa Ayaz

15.3k3939

$endgroup$

add a comment | 

0

0

0

$begingroup$

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

answered Jan 3 '18 at 19:16

Mostafa AyazMostafa Ayaz

15.3k3939

$endgroup$

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

answered Jan 3 '18 at 19:16

Mostafa AyazMostafa Ayaz

15.3k3939

share|cite|improve this answer

share|cite|improve this answer

answered Jan 3 '18 at 19:16

Mostafa AyazMostafa Ayaz

15.3k3939

answered Jan 3 '18 at 19:16

Mostafa AyazMostafa Ayaz

15.3k3939

answered Jan 3 '18 at 19:16

Mostafa AyazMostafa Ayaz

15.3k3939

15.3k3939

add a comment | 

add a comment | 

0

$begingroup$

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

answered Jan 3 '18 at 19:16

TheSimpliFireTheSimpliFire

12.5k62460

$endgroup$

add a comment | 

0

$begingroup$

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

answered Jan 3 '18 at 19:16

TheSimpliFireTheSimpliFire

12.5k62460

$endgroup$

add a comment | 

0

0

0

$begingroup$

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

answered Jan 3 '18 at 19:16

TheSimpliFireTheSimpliFire

12.5k62460

$endgroup$

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

answered Jan 3 '18 at 19:16

TheSimpliFireTheSimpliFire

12.5k62460

share|cite|improve this answer

share|cite|improve this answer

answered Jan 3 '18 at 19:16

TheSimpliFireTheSimpliFire

12.5k62460

answered Jan 3 '18 at 19:16

TheSimpliFireTheSimpliFire

12.5k62460

answered Jan 3 '18 at 19:16

TheSimpliFireTheSimpliFire

12.5k62460

12.5k62460

add a comment | 

add a comment | 

0

$begingroup$

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

answered Jan 3 '18 at 19:20

MastremMastrem

3,79211230

$endgroup$

add a comment | 

0

$begingroup$

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

answered Jan 3 '18 at 19:20

MastremMastrem

3,79211230

$endgroup$

add a comment | 

0

0

0

$begingroup$

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

answered Jan 3 '18 at 19:20

MastremMastrem

3,79211230

$endgroup$

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

answered Jan 3 '18 at 19:20

MastremMastrem

3,79211230

share|cite|improve this answer

share|cite|improve this answer

answered Jan 3 '18 at 19:20

MastremMastrem

3,79211230

answered Jan 3 '18 at 19:20

MastremMastrem

3,79211230

answered Jan 3 '18 at 19:20

MastremMastrem

3,79211230

3,79211230

add a comment | 

add a comment | 

0

$begingroup$

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

answered Jan 3 '18 at 19:29

ss1729ss1729

1,8921723

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Many thanks! :)
  $endgroup$
  – mathsllama
  Jan 3 '18 at 19:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathsllama u could accept appropriate answer.
  $endgroup$
  – ss1729
  Jan 3 '18 at 21:54
  

  
  
  
  

add a comment | 

0

$begingroup$

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

answered Jan 3 '18 at 19:29

ss1729ss1729

1,8921723

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Many thanks! :)
  $endgroup$
  – mathsllama
  Jan 3 '18 at 19:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathsllama u could accept appropriate answer.
  $endgroup$
  – ss1729
  Jan 3 '18 at 21:54
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

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

answered Jan 3 '18 at 19:29

ss1729ss1729

1,8921723

$endgroup$

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

answered Jan 3 '18 at 19:29

ss1729ss1729

1,8921723

share|cite|improve this answer

share|cite|improve this answer

answered Jan 3 '18 at 19:29

ss1729ss1729

1,8921723

answered Jan 3 '18 at 19:29

ss1729ss1729

1,8921723

answered Jan 3 '18 at 19:29

ss1729ss1729

1,8921723

1,8921723

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Many thanks! :)
  $endgroup$
  – mathsllama
  Jan 3 '18 at 19:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathsllama u could accept appropriate answer.
  $endgroup$
  – ss1729
  Jan 3 '18 at 21:54
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Many thanks! :)
  $endgroup$
  – mathsllama
  Jan 3 '18 at 19:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathsllama u could accept appropriate answer.
  $endgroup$
  – ss1729
  Jan 3 '18 at 21:54
  

  
  
  
  

$begingroup$
Many thanks! :)
$endgroup$
– mathsllama
Jan 3 '18 at 19:41

$begingroup$
Many thanks! :)
$endgroup$
– mathsllama
Jan 3 '18 at 19:41

$begingroup$
@mathsllama u could accept appropriate answer.
$endgroup$
– ss1729
Jan 3 '18 at 21:54

$begingroup$
@mathsllama u could accept appropriate answer.
$endgroup$
– ss1729
Jan 3 '18 at 21:54

add a comment | 

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