find the middle term of the expanded form of (2x - 4y^2)^4

[Sham]

find the middle term of the expanded form of (2x - 4y^2)^4

 

[tkhunny]

You can just write them down...

(a+b)^4 will have 5 terms, the pieces of which are:

4!, 0!, 4!,a^4, b^0

4!, 1!, 3!, a^3, b^1

4!, 2!, 2!, a^2, b^2

4!, 3!, 1!, a^1, b^3

4!, 4!, 0!, a^0, b^4

It's ALWAYS the same. Look at those nice, beautiful patterns.

 

[snaper]

(2x-4y^2)^4....

p=4/2=2

middle term....c(4,p).(2x)^2.(4y^2)^2 ... C(4,2)=4!/2!.2!=6

6.4x^2.16y^4

........ + ( 384.x^2.y^4)-........

 

[soroban]

Re: find the middle term of the expanded form of (2x - 4y^2)

Hello, Sham!

Find the middle term of the expanded form of $\displaystyle (2x\,-\,4y^2)^4$

The very worst you can do is multiply it out and look at the middle term . . .


If you're familiar with the Binomial Theorem, we have:

$\displaystyle (2x\,-\,4y)^4\;=\;(2x)^4\,+\,{4\choose3}(2x)^3(-4y^2) \,+\,\underbrace{{4\choose2}(2x)^2(-4y^2)^2}_{\text{middle term}}\,+\,{4\choose1}(2x)(-4y)^3\,+\,(-4y)^4$


Then: \(\displaystyle \:{4\choose2}(2x)^2(-4y^2)^2\;=\;6(4x^2)(16y^4) \;=\;\fbox{384x^2y^4}\)