pascal's triangle

[atlanta213]

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Anyone can explain how pascal's triangle works?

I think i need to understand pascal's triangle to solve below question.

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

Thanks!

 

[klutzy1987]

Anyone can explain how pascal's triangle works?

I think i need to understand pascal's triangle to solve below question.

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

Thanks!

Yes you need pascals triangle for this.

http://en.wikipedia.org/wiki/Pascals_triangle

It is too complicated to explain from scratch, read this and then post any questions that you have.

 

[admap]

I got what the Pascal triangle is and how to construct it, just not sure how to use it for the question Op has posted...
Please explain...Thanks

 

[UCfan]

Dear Atlanta213,
First you need construct the table and understand it. You have (a-b)^5.
It can be useful for you to understand this example:

(a-b)^5=1(a^5)-5(a^4)(b)+10(a^3)(b^2)-10(a^2)(b^3)+5(a)(b^4)-1(b^5)

Right now let's look at your problem:
(3x-2y)^5= 1((3x)^5)-5((3x)^4)(2y)+10((3x)^3)((2y)^2)-10((3x)^2)((2y)^3)+
5(3x)((2y)^4)-1((2y)^5)

Little bit simplification gives:
=243(x^5)-5(81)(2)(x^4)👍+10(27)(4)(x^3)(y^2)-10(9)(8)(x^2)(y^3)
+5(3)(16)(x)(y^2)-1(32)(y^5)

More simplification gives you:
=243(x^5)-810(x^4)+1080(x^3)(y^2)-720(x^2)(y^3)+240(x)(y^2)-32(y^5) (Check my math maybe I made a multiplication mistake 🙂)
Hopefully it helps!

Sincerely!

 

[Streetwolf]

It gives you the coefficient for the term. So (5 C 1) gives you the coefficient for the 2nd term (because (5 C 0) is the first term). The rest of the term is the x and the y. Remember that the first term is x^5*y^0 (which equals just x^5). The second term is x^4*y^1. The third term is x^3*y^2. See the pattern? This works for any exponent in the binomial. The x term exponent starts at the exponent in the binomial while the y term exponent starts at 0. Each successive term, the x term exponent decreases by 1 while the y term exponent increases by 1, until the last term is x^0 * y^n.

Remember in this problem that x really is 3x and y really is -2y. So when you take your second term which should be 5x^4*y^1, it is really 5(3x)^4(-2y)^1.

 

[Thundercatz]

If I see this on the test, I am hitting the skip button! HAHA