Problem With Binomial Coefficients

Hello,

I am trying to solve the following problem:

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

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

Can anyone help with this question?

Re: Problem With Binomial Coefficients

Do you mean:

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

Re: Problem With Binomial Coefficients

Quote:

Originally Posted by

**MarkFL2** Do you mean:

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

Correct! Sorry about that!

Re: Problem With Binomial Coefficients

Okay, the binomial theorem tells us the general term in this expansion will be:

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

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

Re: Problem With Binomial Coefficients

Quote:

Originally Posted by

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

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

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?

Re: Problem With Binomial Coefficients

Yes, correct on both counts! :D

So, what would the value of that term be?

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 $\displaystyle {8 \choose 2} = 28$?

Thanks! :D

Re: Problem With Binomial Coefficients