find the middle term of the expanded form of (2x - 4y^2)^4
- Thread starter Sham
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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.
(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.
Re: find the middle term of the expanded form of (2x - 4y^2)
Hello, Sham!
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: \(\displaystyle \:{4\choose2}(2x)^2(-4y^2)^2\;=\;6(4x^2)(16y^4) \;=\;\fbox{384x^2y^4}\)
Hello, Sham!
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: \(\displaystyle \:{4\choose2}(2x)^2(-4y^2)^2\;=\;6(4x^2)(16y^4) \;=\;\fbox{384x^2y^4}\)