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May 25, 2019, 10:53:35 am

Author Topic: Pascal's triangle  (Read 2912 times)  Share 

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Collin Li

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Pascal's triangle
« Reply #15 on: November 01, 2007, 12:06:14 am »
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Quote from: "Toothpick"
(2x+3)^7 ..

so the coefficient of x^4 is

(7 C 4) (2x)^4 3^(3)

15120?


Ehh, yeah, but only by luck:

x^7, x^6, x^5, x^4

x^4 is the 4th term, which means we're looking at column 4, so r+1=4 => r=3.

7C3 is what you needed, but 7C4 = 7C3 because of the identity: nCk = nC(n-k)

Try again if I had asked for the x^5 term?

Toothpaste

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Pascal's triangle
« Reply #16 on: November 01, 2007, 12:15:51 am »
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7-r = 5   (is this valid? It works lol)
r=2

but yeah it's the 3rd term so r +1 = 3 , r=2

coefficient of x^5 would be

(7 C 2) (2x)^5 3^(2)

6048

Toothpaste

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« Reply #17 on: November 01, 2007, 12:31:43 am »
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Yeah, it's clear now.


(2x+3)^7
= (2x)^7 + 7(2x)^6 3 + 21 (2x)^5 3^2 ... etc

:D

Collin Li

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« Reply #18 on: November 01, 2007, 12:46:59 am »
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Quote from: "Toothpick"
7-r = 5   (is this valid? It works lol)
r=2

but yeah it's the 3rd term so r +1 = 3 , r=2

coefficient of x^5 would be

(7 C 2) (2x)^5 3^(2)

6048


Yeah, using n-r = k'th term works fine as well. I think that's the proper definition anyway, my r+1 = k'th term happens to work as well, though :P