# Thread: Problem With Binomial Coefficients

1. ## 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?

2. ## Re: Problem With Binomial Coefficients

Do you mean:

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

3. ## Re: Problem With Binomial Coefficients

Originally Posted by MarkFL2
Do you mean:

$\displaystyle \left(x+\left(\frac{1}{x} \right)^3 \right)^8$ ?
Correct! Sorry about that!

4. ## 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?

5. ## Re: Problem With Binomial Coefficients

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?

6. ## Re: Problem With Binomial Coefficients

Yes, correct on both counts!

So, what would the value of that term be?

7. ## Re: Problem With Binomial Coefficients

Originally Posted by MarkFL2
Yes, correct on both counts!

So, what would the value of that term be?
It would be $\displaystyle {8 \choose 2} = 28$?

Thanks!

8. ## Re: Problem With Binomial Coefficients

Yes, correct again!