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# Try something new - Expansion challenge.

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Find the coefficient of  \( x^2y^3z\)     in the expansion of    \((4x+y-2z)^6\)

Heureka shows you how to do it here:

https://web2.0calc.com/questions/binomial-theroem

Perhaps you could find all the terms this way although there are an aweful lot of them.

Sep 17, 2018

### 7+0 Answers

#1
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Find the coefficient of x^2y^3z in the expansion of (4x + y - 2z)^6
The coefficient is calculated as follows:
[6!(original exponent) / 2!.3!.1!(exponents of the desired term)] X [4x(First term of original trinomial)]^(desired power(2)) X [y(Second term of original trinomial)]^(desired power(3)) X [2z(Third term of original trinomial)]^(desired power(1))
6! /2!.3!.1!  * (4x)^2 * (y^3) * (2z)=60 * 16x^2 * y^3 *2z=1920x^2y^3z

Sep 17, 2018
edited by Guest  Sep 17, 2018
#2
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[Note: The number of terms of this expansion is calculated as follows:( 6(The exponent of the trinomial)) + number of terms - 1 nCr the exponent of the trinomial =6 + 3 -1 nCr 6 =8 nCr 6 =28 terms]

Thanks to my CAS system, here are ALL the 28 terms and their coefficients as calculated above:
4096 x^6 + 6144 x^5 y - 12288 x^5 z + 3840 x^4 y^2 - 15360 x^4 y z + 15360 x^4 z^2 + 1280 x^3 y^3 - 7680 x^3 y^2 z + 15360 x^3 y z^2 - 10240 x^3 z^3 + 240 x^2 y^4 - 1920 x^2 y^3 z + 5760 x^2 y^2 z^2 - 7680 x^2 y z^3 + 3840 x^2 z^4 + 24 x y^5 - 240 x y^4 z + 960 x y^3 z^2 - 1920 x y^2 z^3 + 1920 x y z^4 - 768 x z^5 + y^6 - 12 y^5 z + 60 y^4 z^2 - 160 y^3 z^3 + 240 y^2 z^4 - 192 y z^5 + 64 z^6
(28 terms)

Sep 17, 2018
#3
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Thanks guest,

It is great to see that my post sparked some interest. :)

The first one is great.

You did make one small error though. (It is a very common error all through mathematics)

Can you find it?

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The second one is also good. I haver not checked it I am just impressed that you did it.  I assume it is correct since you did it with a program.

However I do not understand your formula.  How can 6+3- a positive number =28

mmmm

Oh maybe my detective work has worked.... do you mean  ???

nCr

where  n=6+3-1          [Exponent of polynomial + number of terms - 1]

and r = 6                     [exponent of polynomial]

So that is       8C6 = 28       Bingo So for a nomal cubic binomial  (a+b)^3    There would be (3+2-1)C3 = 4C3 = 4 terms.

Das ist Ausgezeichnet! Sep 17, 2018
#4
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Hi Melody: Thanks for the compliment. Is it the (-) sign before the coefficient?

Number of terms = The exponent of the trinomial =6 + Number of terms= 3 - 1=8C(the exponent of the trinomial) =6. Or, as you calculated: 8C6 =28.

Sep 17, 2018
#5
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Yes it is       (-2z)^1                Because the terms are    4x+y+(-2z)

I am very pleased that you are so interested in mathematics and that you learned from this challenge. It would be very nice if you worked on your presention.

Yours all runs together so it is very difficult to make sense of. Plus you do not use brackets when they are needed.

For most people maths is really hard. When they see a big block of letters and numbers and symbols they just turn off.

That is not what you want. Melody  Sep 17, 2018
#6
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Thanks again. I did think of both mistakes but was too lazy, especially the calculation of the number of terms of the expansion. I thought that was "known" to most people interested in Math! Now, I know better!. I didn't know you spoke German!!!.

Sep 17, 2018
#7
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I don't. I am just trying to learn. LOL

Melody  Sep 17, 2018