binomial

**JQ2009** · November 1st 2009, 05:16 AM

How would you find the term x^2*y*z^5 in the expansion of (x-2y+z)^8? First you have to expand ((x-2y)+z)^8.

I understand how to find the coefficient but I don't really understand how to find the term.

**tonio** · November 1st 2009, 06:39 AM

Originally Posted by JQ2009

How would you find the term x^2*y*z^5 in the expansion of (x-2y+z)^8? First you have to expand ((x-2y)+z)^8.

I understand how to find the coefficient but I don't really understand how to find the term.

Try to do it in two steps: as $(a+b)^n=\sum\limits_{k=0}^n\binom {n}{k}a^kb^{n-k}$ , we get:

$((x-2y)+z)^8=\sum\limits_{k=0}^n\binom {8}{k}(x-2y)^kz^{8-k}$

So you must choose k = 3 above to get $z^5$ : $\binom {8}{3}(x-2y)^3z^5$ , and now you must choose the correct power in $(x-2y)^3$ to finally get $x^2yz^5$

Tonio

**JQ2009** · November 1st 2009, 07:20 AM

What do you mean with choose the correct power?

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