MHB Constant term in a binomial expansion

[Petrus]

Decide constant term in $$\left(3\cdot x^3+\left(\frac{-4}{x} \right) \right)^{20}$$.

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

 

[chisigma]

Re: constant term

Decide constant term in https://webwork.math.su.se/webwork2_files/tmp/equations/67/172c674f7e20bbdefabddc24741bac1.png.

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]

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]

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]

Re: constant term

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

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

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

 

[Petrus]

Re: constant term

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

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

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]

Re: constant term in a binomial expansion

Yes, you are absolutely correct! (Yes)

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

 

[Petrus]

Re: constant term

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

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

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]

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:

$$C=-3^5\cdot4^{15}{20 \choose 15}$$

or even

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

Do you know the identity $${n \choose r}={n \choose n-r}$$?

 

[Petrus]

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:

$$C=-3^5\cdot4^{15}{20 \choose 15}$$

or even

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

Do you know the identity $${n \choose r}={n \choose n-r}$$?

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]

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]

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:

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

 

[Petrus]

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]

A good way is to use the definition:

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

which means the identity is:

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

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

 

[MarkFL]

Hello Petrus,

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

Spoiler

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

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

First, we may write:

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

and so our general term may be written:

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

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

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

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

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

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

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

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

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

Simplifying the exponents, we find:

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

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

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

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

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

 

[Petrus]

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:

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

Thanks Mark!
I succed to solve it after a lot thinking and attempt!
This kind of problem should be a lot of cause its not always I can use exponent rules and at end you forget the rules!:/