# Problem With Binomial Coefficients

• Jan 16th 2013, 09:35 PM
ianm
Problem With Binomial Coefficients
Hello,

I am trying to solve the following problem:

"Find the term independent of x in $(x+\frac{1}{x})^3)^8$."

I think this formula applies, but I am not sure where to begin: $t_{k+1}={n \choose k}x^{n-k}y^k$.

Can anyone help with this question?
• Jan 16th 2013, 10:02 PM
MarkFL
Re: Problem With Binomial Coefficients
Do you mean:

$\left(x+\left(\frac{1}{x} \right)^3 \right)^8$ ?
• Jan 16th 2013, 10:20 PM
ianm
Re: Problem With Binomial Coefficients
Quote:

Originally Posted by MarkFL2
Do you mean:

$\left(x+\left(\frac{1}{x} \right)^3 \right)^8$ ?

• Jan 16th 2013, 10:33 PM
MarkFL
Re: Problem With Binomial Coefficients
Okay, the binomial theorem tells us the general term in this expansion will be:

${8 \choose k}x^{8-k}\left(\left(\frac{1}{x} \right)^3 \right)^k={8 \choose k}x^{8-k}x^{-3k}={8 \choose k}x^{8-4k}$

Now, in order for this term to be independent of x, to what must the exponent on x be equal?
• Jan 16th 2013, 10:40 PM
ianm
Re: Problem With Binomial Coefficients
Quote:

Originally Posted by MarkFL2
Okay, the binomial theorem tells us the general term in this expansion will be:

${8 \choose k}x^{8-k}\left(\left(\frac{1}{x} \right)^3 \right)^k={8 \choose k}x^{8-k}x^{-3k}={8 \choose k}x^{8-4k}$

Now, in order for this term to be independent of x, to what must the exponent on x be equal?

0, correct? Therefore, would k be 2?
• Jan 16th 2013, 10:44 PM
MarkFL
Re: Problem With Binomial Coefficients
Yes, correct on both counts! :D

So, what would the value of that term be?
• Jan 16th 2013, 10:46 PM
ianm
Re: Problem With Binomial Coefficients
Quote:

Originally Posted by MarkFL2
Yes, correct on both counts! :D

So, what would the value of that term be?

It would be ${8 \choose 2} = 28$?

Thanks! :D
• Jan 16th 2013, 10:52 PM
MarkFL
Re: Problem With Binomial Coefficients
Yes, correct again! :D