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# Expansive Expansion: Newsletter Challenge (December 2013)

Posted December 01, 2013 12:00 AM
Pathfinder Tags: challenge questions

This month's Challenge Question: Specs & Techs from GlobalSpec:

A binomial raised to the 3rd power and then algebraically expanded has binomial coefficients of 1, 3, 3, 1. What are the binomial coefficients when a binomial is raised to the 16th power?

Pascal's Triangle makes this an easy problem to solve. Simply use the bottom row below to obtain the coefficients for each term. For trinomial expansions there is a similar tool called Pascal's pyramid.

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#1

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/01/2013 1:02 AM

1, 16, 120, 560, 1820, 4368, 8008, 11440, 12870, 11440, 8008, 4368, 1820, 560, 120, 16, 1

Never memorize something that you can look up.
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#2

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/01/2013 3:56 AM

If you don't want to look it up, just do a Pascal's Triangle.

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#3

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/01/2013 1:18 PM

Yup.

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#9

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/02/2013 2:05 PM

Ditto.

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#4

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/01/2013 2:54 PM

Since the general form is (ax + by)n, the coefficients of the expanded general form of the binomial to the 16th power would be:

1a16, 16a15b, 120a14b2 560a13b3, 1820a12b4, 4368a11b5, 8008a10b6, 11440a9b7, 12870a8b8, 11440a7b9, 8008a6b10, 4368a5b11, 1820a4b12, 560a3b13, 120a2b14, 16ab15, 1b16

/Another exciting challenge.

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#5

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/01/2013 3:28 PM

Yup.

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#6

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/02/2013 4:21 AM

This time they can publish the answer on the first, rather than last, Tuesday of the month.

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#7

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/02/2013 9:58 AM

As doorman would say: yup.

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#8

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/02/2013 12:14 PM

Yup.

I remember this being ladled out in high school physics or trig. Is this still the case?

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#10

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/03/2013 12:46 AM

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#11

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/03/2013 11:07 AM

In case you didn't guess, what SolarEagle showed you here was the aforementioned "Pascal's Triangle" (or at least the top 17 rows...)

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#12

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/03/2013 2:17 PM

Generally, any coefficient is n!/(m!*(n-m)!) where n is the binomial exponent and m is the ordinate of the coefficients of the expansion working from left to right (or right to left - gotta love symmetry)

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#13

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/03/2013 5:46 PM

That is why I leave the higher algebra to those who are complete whizzed (Ghengi even) at algebra. I hate to admit the answer did not automatically roll off the tip of my tongue, and as usual, am the last dog to the bowl anyway.

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#14

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/03/2013 6:06 PM

1,16,120,560,1820,4368,8008,11440,12870,11440,8008,4368,8008,4368,1820,560,

120,16,1. I am glad to see that my notes on Pascals triangle from 25 years ago are correct.

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#15

### Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/04/2013 12:21 AM

the binominal coeffizients ar3:

1 16 120 560 1820 4368 8008 11440 12870 12870 11440 8008 4368 1820 560 120 16 1

or take a pohl

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