quantitative reasoning destroyer 149 (2011)

[faye1891]

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Hey guys,

I just need help clarifying this question/making it easier and faster to figure out.

149. What is the product of the second and third term of the expansion (3x-2y)^5?


The answer is -874,800x^7y^3t
They used Pascal's triangle to figure it out but I didn't really understand the explanation.
Is there a way I can do this problem FAST without having to actually expand it out which could take forever? How do I use Pascal's Triangle for this problem (I'm not very familiar with it)?

Thanks!!!

 

[lr2014]

I doubt a question like that would be in the DAT, their questions have to be solved in under 1.5 min at most I think.

 

[kl352]

first of all, i'm quite certain the real dat wil never ask a question as complicated or time consuming. if anything, they would probably ask for a term of something to like 3rd power.

i think first thing to understand is to forget the coefficients, understand that there will be 6 terms, x^5, x^4 * y, x^3 y^2, x^2 y^3, x y^4, and y^5. number of terms just add one to the power. so we look at the second and third term, multiply them together and that's where you get (x^4 * y) * (x^3 * y^2) = x^7 * y^3. now look at the answer choice and see if any of the terms have the variables with those powers. if they ask anything like this, that's probably all the calculation you have to do because it is kind of a bitch to calculate the coefficient.

for the sake of entirely understanding the problem, to calculate the coefficient...
(x+y)^1 = x + y coefficents: 1,1
(x+y)^2 = x^2 + 2xy + y^2 coefficients: 1,2,1
(x+y)^3 = x^3 + 3x^2 y + 3xy^2 + y^3 coefficients: 1,3,3,1

do you see a pattern? essentially for the 1s on each side, you just move it down, and then the inner terms, you just add consecutive integers. so for 1,2,1. move the 1s on the sides, and add 1+2 and 2+1 so you get 1,3,3,1. the next one will be 1 4 6 4 1 (1, 1+3, 3+3, 3+1, 1). remarkably, these are the coefficients for (x+y) to whatever power! so now that we have 6 terms, lets get the next row for the triangle.

1 4 6 4 1
1 5 10 10 5 1

so coefficient for the second term is formally 5 * (3x)^4 * (-2y) ^ 1
for the third term is 10 * (3x)^3 * (-2y)^2

put it together

5 * 10 * 3^7 * (-2)^3 *x^7 * y^3 = -874800

hope that helps. just know that there's like no way they'll ever ask anything as time consuming as this

 

[ballyhoo]

1 4 6 4 1
1 5 10 10 5 1

so coefficient for the second term is formally 5 * (3x)^4 * (-2y) ^ 1
for the third term is 10 * (3x)^3 * (-2y)^2

put it together

5 * 10 * 3^7 * (-2)^3 *x^7 * y^3 = -874800

hope that helps. just know that there's like no way they'll ever ask anything as time consuming as this

Wait...how did you get the 1 4 6 4 1? How do you get the inside coefficient? For example, in your explanation, you say the coefficient is 5 * (3x)^4 <-- just for x, how do you get the 3?

 

[kl352]

the row before with 4 terms if 1 3 3 1

1 3 3 1
1 (1+3) (3+3) (3+1) 1 -> 1 4 6 4 1 always add consecutive terms

the second term 5* (3x)^4 * (-2y)^1 i got the 3 because the problem states (3x-2y)^5 so when doing the powers, you need to apply them to (3x) and (-2y)