Problem With Binomial Coefficients

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January 16th 2013, 08:35 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?
January 16th 2013, 09:02 PM
MarkFL

Re: Problem With Binomial Coefficients

Do you mean:

$\left(x+\left(\frac{1}{x} \right)^3 \right)^8$ ?
January 16th 2013, 09:20 PM
ianm

Re: Problem With Binomial Coefficients

Quote:

Originally Posted by MarkFL2

Do you mean:

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

Correct! Sorry about that!
January 16th 2013, 09:33 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?
January 16th 2013, 09:40 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?
January 16th 2013, 09:44 PM
MarkFL

Re: Problem With Binomial Coefficients

Yes, correct on both counts! :D

So, what would the value of that term be?
January 16th 2013, 09:46 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
January 16th 2013, 09:52 PM
MarkFL

Re: Problem With Binomial Coefficients

Yes, correct again! :D

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