# Thread: Find the coefficient (Precalc II)

1. ## Find the coefficient (Precalc II)

Could someone please help solve the following:

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!

ETA: Answer = 294912

5. Originally Posted by Plato
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 ${9 \choose 8}\left( {\frac{x}{2}} \right)^1 \left( { - 4y} \right)^8$
Plato, do you mind elaborating as to how your expansion method differs from galactus'? Would this be an acceptable method (especially when showing your work)? Also, it seems as if the 'y' is ignored in '-4y'. Or is just y=1 and it's multiplied by the 4. Can you explain it or is just one of those things that you just do and don't think about ? Thanks!