Question

Find the coefficient of x^4 in the expansion of (2x-1)^15, using binomial theorem

Answer 1

Hey! Do you know how to use Pascal's triangle?

Answer 2

Yes..but I need to use the other method

Answer 3

Using the Binomial Theorem, our Kth term will look like this,
\[\Large\rm \left(\begin{matrix}15 \\ k\end{matrix}\right)(2x)^{15-k}(-1)^{k}\]

So how do we get a 4th power on our x?
What should our k value be so we end up with a power of 4?

Answer 4

k=11

Answer 5

\[\Large\rm \left(\begin{matrix}15 \\ 11\end{matrix}\right)(2x)^{15-11}(-1)^{11}\]Ok good.

Answer 6

\[\Large\rm \left(\begin{matrix}15 \\ 11\end{matrix}\right)(2x)^{4}(-1)^{11}\]

Answer 7

We've got a ton of numbers floating around, so we'll have to be very careful.

Answer 8

(-1)^{11} = ?

Answer 9

Negative to an odd power.
What does that simplify to? :d

Answer 10

negative..

Answer 11

-1

Answer 12

Ok good, let's bring it to the front.
\[\Large\rm -\left(\begin{matrix}15 \\ 11\end{matrix}\right)(2x)^{4}\]

Answer 13

How bout the 2x?
We have to apply the 4th power to `both` the 2 and the x.

Answer 14

\[\Large\rm -\left(\begin{matrix}15 \\ 11\end{matrix}\right)2^4x^4\]

Answer 15

So we'll bring that to the front as well, yes?

Answer 16

\[\Large\rm -16\left(\begin{matrix}15 \\ 11\end{matrix}\right)x^4\]

Answer 17

So then we need to deal with this thing:
\[\Large\rm \left(\begin{matrix}n \\ k\end{matrix}\right)=\frac{n!}{k!(n-k)!}\]Applying it to our problem:\[\Large\rm \left(\begin{matrix}15 \\ 11\end{matrix}\right)=\frac{15!}{11!(15-11)!}\]

Answer 18

ok..

Answer 19

Those factorials are really big nasty numbers.
So you'll want to try and cancel things out before expanding out the numbers.

Answer 20

Here is what I would do....

Answer 21

what happen to the negative sign

Answer 22

The -16? We're not looking at that part right now.

Answer 23

Is that the negative that you're asking about?

Answer 24

okay

Answer 25

We're going to rewrite our numerator like this:
\[\Large\rm 15!=15\cdot14\cdot13\cdot12\cdot11!\]Does it make sense that they are equal?

Answer 26

yes coz 15factorial

Answer 27

\[\Large\rm \left(\begin{matrix}15 \\ 11\end{matrix}\right)=\frac{15\cdot14\cdot13\cdot12\cdot11!}{11!(4)!}\]

Answer 28

We have a nice cancellation from there,\[\Large\rm \left(\begin{matrix}15 \\ 11\end{matrix}\right)=\frac{15\cdot14\cdot13\cdot12\cdot\cancel{11!}}{\cancel{11!}(4)!}\]

Answer 29

kk..

Answer 30

Then maybe just put it into a calculator from there.
What value do you get?

Answer 31

1365

Answer 32

\[\Large\rm -16\left(\begin{matrix}15 \\ 11\end{matrix}\right)x^4\]Ok great!\[\Large\rm -16\left(1365\right)x^4\]

Answer 33

and then just multiply to simplify it the rest of the way.

Confused at any parts?
You were droppin' a lot of dots on me (ok.... ) so I thought maybe you got stuck somewhere lol

Answer 34

last part..what happens to x^4
-21840x^4?

Answer 35

Ok great! So you've successfully found the coefficient for x^4 in that expansion.\[\Large\rm -21840\]That's your answer.

Answer 36

Alrite, thanks for the clear explanation @zepdrix

Answer 37

np c: