# Math Help - Find the coefficient (Precalc II)

1. ## Find the coefficient (Precalc II)

Find the coefficient of the term containing $y^8$ in the expansion of:
$[(x/2) - 4y]^9$

Can you please list the steps/formula and how this should be solved? TIA!

2. Use the expansion:

$(a+b)^{n}=a^{n}+na^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}b^{2}+$ $..........+\frac{n(n-1)(n-2).....(n-k+1)}{k!}a^{n-k}b^{k}+............+nab^{n-1}+b^{n}$

Just list them out: $a=\frac{x}{2}, \;\ b=-4y, \;\ n=9$

Here's the first few:

$(\frac{x}{2})^{9}+9(\frac{x}{2})^{8}(-4y)+\frac{9(8)}{2!}(\frac{x}{2})^{7}(-4y)^{2}+...............+b^{9}$

$=\frac{x^{9}}{512}-\frac{9}{64}x^{8}y+\frac{9}{2}x^{7}y^{2}+......... .... -262144y^{9}$

List them out until you find the $y^{8}$ term.

Do you see he pattern?. If so, you can find it without listing them all out. But it would be good to allow you to see what's going on.

3. Each term in the expansion of $\left( {a + b} \right)^N$ looks like ${N \choose k}a^k b^{N - k}$.

Now in your problem $a = \frac{x}{2}$, $b = - 4y$ and $N = 9$.
But you want the term containing $y^8$ so ${9 \choose 8}\left( {\frac{x}{2}} \right)^1 \left( { - 4y} \right)^8$

4. Originally Posted by galactus
Use the expansion:

$(a+b)^{n}=a^{n}+na^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}b^{2}+$ $..........+\frac{n(n-1)(n-2).....(n-k+1)}{k!}a^{n-k}b^{k}+............+nab^{n-1}+b^{n}$

Just list them out: $a=\frac{x}{2}, \;\ b=-4y, \;\ n=9$

Here's the first few:

$(\frac{x}{2})^{9}+9(\frac{x}{2})^{8}(-4y)+\frac{9(8)}{2!}(\frac{x}{2})^{7}(-4y)^{2}+...............+b^{9}$

$=\frac{x^{9}}{512}-\frac{9}{64}x^{8}y+\frac{9}{2}x^{7}y^{2}+......... .... -262144y^{9}$

List them out until you find the $y^{8}$ term.

Do you see he pattern?. If so, you can find it without listing them all out. But it would be good to allow you to see what's going on.
I see the pattern, but WOW! I'm going to do it on paper and post the final answer. Thanks!

Each term in the expansion of $\left( {a + b} \right)^N$ looks like ${N \choose k}a^k b^{N - k}$.
Now in your problem $a = \frac{x}{2}$, $b = - 4y$ and $N = 9$.
But you want the term containing $y^8$ so ${9 \choose 8}\left( {\frac{x}{2}} \right)^1 \left( { - 4y} \right)^8$