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Constant term in a binomial expansion

Petrus

Well-known member
Feb 21, 2013
739
Decide constant term in \(\displaystyle \left(3\cdot x^3+\left(\frac{-4}{x} \right) \right)^{20}\).

I have problem with this one, I can't find any example about this one in my book, any advice would be great:)
 
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chisigma

Well-known member
Feb 13, 2012
1,704
Re: constant term

Decide constant term in
.

I have problem with this one, I can't find any exemple about this one in my book, any advice would be great:)
If You apply the bynomial sum...

$\displaystyle (a+b)^{n} = \sum_{k=0}^{n} \binom{n}{k}\ a^{k}\ b^{n-k}$ (1)

... with $a=3\ x^{3}$, $a=- \frac{4}{x}$ and $n=20$ the constant term is for $\displaystyle 3\ k = 20 - k \implies k=5$ so that the connstant term is...


$\displaystyle c= \binom{20}{5}\ 3^{5}\ (-4)^{15}$ (2)


Kind regards


$\chi$ $\sigma$
 

Petrus

Well-known member
Feb 21, 2013
739
Re: constant term

If You apply the bynomial sum...

$\displaystyle (a+b)^{n} = \sum_{k=0}^{n} \binom{n}{k}\ a^{k}\ b^{n-k}$ (1)

... with $a=3\ x^{3}$, $a=- \frac{4}{x}$ and $n=20$ the constant term is for $\displaystyle 3\ k = 20 - k \implies k=5$ so that the connstant term is...


$\displaystyle c= \binom{20}{5}\ 3^{5}\ (-4)^{15}$ (2)


Kind regards

$\chi$ $\sigma$
Thanks!
you got a typo :p $a=- \frac{4}{x}$ it should be 'b' not 'a'. Unfortently this was my homework (I am glad that you helped me ) but next time just give me tips,advice :p I am suposed to solve my homework not you :p
 

Petrus

Well-known member
Feb 21, 2013
739
Re: constant term

If You apply the bynomial sum...

$\displaystyle (a+b)^{n} = \sum_{k=0}^{n} \binom{n}{k}\ a^{k}\ b^{n-k}$ (1)

... with $a=3\ x^{3}$, $a=- \frac{4}{x}$ and $n=20$ the constant term is for $\displaystyle 3\ k = 20 - k \implies k=5$ so that the connstant term is...


$\displaystyle c= \binom{20}{5}\ 3^{5}\ (-4)^{15}$ (2)


Kind regards


$\chi$ $\sigma$
Hello,
After reading this problem and try solve it I get hard understand why you use as x=1 and how do you get this $\displaystyle 3\ k = 20 - k \implies k=5$?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: constant term

The general term, as given by the binomial theorem, is:

\(\displaystyle {20 \choose k}\left(3x^3 \right)^{20-k}\left(-\frac{4}{x} \right)^k=3^{20-k}(-4)^k{20 \choose k}x^{3(20-k)}x^{-k}=3^{20-k}(-4)^k{20 \choose k}x^{4(15-k)}\)

What must the exponent on $x$ be in order for the term to be a constant?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: constant term

The general term, as given by the binomial theorem, is:

\(\displaystyle {20 \choose k}\left(3x^3 \right)^{20-k}\left(-\frac{4}{x} \right)^k=3^{20-k}(-4)^k{20 \choose k}x^{3(20-k)}x^{-k}=3^{20-k}(-4)^k{20 \choose k}x^{4(15-k)}\)

What must the exponent on $x$ be in order for the term to be a constant?
$x^{4(15-k)}$ we want it to be power up to 0 so it becomes 1, so $15-k=0 <=> k=15$
I am correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: constant term in a binomial expansion

Yes, you are absolutely correct! (Yes)

So, letting $k=15$, what is the constant term?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: constant term

The general term, as given by the binomial theorem, is:

\(\displaystyle {20 \choose k}\left(3x^3 \right)^{20-k}\left(-\frac{4}{x} \right)^k=3^{20-k}(-4)^k{20 \choose k}x^{3(20-k)}x^{-k}=3^{20-k}(-4)^k{20 \choose k}x^{4(15-k)}\)

What must the exponent on $x$ be in order for the term to be a constant?
Replying to this for some latex mall:)
$C=\frac{20!}{15!(20-15)!}3^5(-4)^{15}$
is this correct way to answer this problem?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: constant term in a binomial expansion

That's one way to express the constant term. It is a very large number, so I think I would leave it as:

\(\displaystyle C=-3^5\cdot4^{15}{20 \choose 15}\)

or even

\(\displaystyle C=-\left(3\cdot4^3 \right)^5{20 \choose 5}\)

Do you know the identity \(\displaystyle {n \choose r}={n \choose n-r}\)?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: constant term in a binomial expansion

That's one way to express the constant term. It is a very large number, so I think I would leave it as:

\(\displaystyle C=-3^5\cdot4^{15}{20 \choose 15}\)

or even

\(\displaystyle C=-\left(3\cdot4^3 \right)^5{20 \choose 5}\)

Do you know the identity \(\displaystyle {n \choose r}={n \choose n-r}\)?
That last one i did not know, why would i like to write it on that form? There is so many diffrent form to write but in exam how would you answer?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: constant term in a binomial expansion

You are right that there are many forms to write this constant, and unless you are directed to write it a certain way, I suppose it is up to you how you choose to express the result.

I would choose the last form I gave simply because it's just ever so slightly more compact than the first form I gave.

In the form you gave, you would certainly want to simplify $(20-15)!$ to $5!$.

As a follow-up, can you demonstrate that the combinatorial identity I cited is true?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: constant term in a binomial expansion

Petrus has asked that I give him another similar problem for practice that he can try to solve. I decided to post it here for the benefit of others perhaps reading this topic who are wanting help with this kind of problem:

Find the constant term in the expansion of:

\(\displaystyle x^3\left(\frac{x^3}{y^2}-\frac{3y}{x^2} \right)^9\)
 

Petrus

Well-known member
Feb 21, 2013
739
Re: constant term in a binomial expansion

You are right that there are many forms to write this constant, and unless you are directed to write it a certain way, I suppose it is up to you how you choose to express the result.

I would choose the last form I gave simply because it's just ever so slightly more compact than the first form I gave.

In the form you gave, you would certainly want to simplify $(20-15)!$ to $5!$.

As a follow-up, can you demonstrate that the combinatorial identity I cited is true?
Hello Mark,
I can't see why it's true
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
A good way is to use the definition:

\(\displaystyle {n \choose r}\equiv\frac{n!}{r!(n-r)!}\)

which means the identity is:

\(\displaystyle \frac{n!}{r!(n-r)!}=\frac{n!}{(n-r)!(n-(n-r))!}\)

Can you see why this is true? Can you relate this identity to Pascal's triangle?
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Petrus,

A solution to the problem I posted is hidden below, so that you may check your work:

The binomial theorem tells us the general term in the expansion is:

\(\displaystyle x^3{9 \choose k}\left(\frac{x^3}{y^2} \right)^{9-k}\left(-\frac{3y}{x^2} \right)^k\)

First, we may write:

\(\displaystyle \left(-\frac{3y}{x^2} \right)^k=(-3)^k\left(\frac{y}{x^2} \right)^k\)

and so our general term may be written:

\(\displaystyle x^3(-3)^k{9 \choose k}\left(\frac{x^3}{y^2} \right)^{9-k}\left(\frac{y}{x^2} \right)^k\)

Next using the property of exponents \(\displaystyle \left(\frac{a}{b} \right)^c=\frac{a^c}{b^c}\) we may write the term as:

\(\displaystyle (-3)^k{9 \choose k}x^3\frac{\left(x^3 \right)^{9-k}}{\left(y^2 \right)^{9-k}}\cdot\frac{y^k}{\left(x^2 \right)^k}\)

Now we may use the property of exponents \(\displaystyle \left(a^b \right)^c=a^{bc}\) to write:

\(\displaystyle (-3)^k{9 \choose k}x^3\cdot\frac{x^{3(9-k)}}{y^{2(9-k)}}\cdot\frac{y^k}{x^{2k}}
\)

Using the property of exponents \(\displaystyle \frac{1}{a^b}=a^{-b}\) we may write:

\(\displaystyle (-3)^k{9 \choose k}x^3\cdot x^{3(9-k)}\cdot y^{-2(9-k)}\cdot y^k\cdot x^{-2k}\)

Using the property of exponents \(\displaystyle a^b\cdot a^c=a^{b+c}\) we may write:

\(\displaystyle (-3)^k{9 \choose k}x^{3+3(9-k)-2k}\cdot y^{-2(9-k)+k}\)

Simplifying the exponents, we find:

\(\displaystyle (-3)^k{9 \choose k}x^{3+27-3k-2k}\cdot y^{-18+2k+k}\)

\(\displaystyle (-3)^k{9 \choose k}x^{30-5k}\cdot y^{3k-18}\)

\(\displaystyle (-3)^k{9 \choose k}x^{5(6-k)}\cdot y^{3(k-6)}\)

We find then, that for $k=6$, the term is constant, and given by:

\(\displaystyle (-3)^6{9 \choose 6}=729\cdot84=61236\)
 

Petrus

Well-known member
Feb 21, 2013
739
Re: constant term in a binomial expansion

Petrus has asked that I give him another similar problem for practice that he can try to solve. I decided to post it here for the benefit of others perhaps reading this topic who are wanting help with this kind of problem:

Find the constant term in the expansion of:

\(\displaystyle x^3\left(\frac{x^3}{y^2}-\frac{3y}{x^2} \right)^9\)
Thanks Mark!
I succed to solve it after alot thinking and attempt!
This kind of problem should be alot of cause its not always I can use exponent rules and at end you forget the rules!:/