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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?



And the answer is:

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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Doorman
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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
In reply to #2

Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/01/2013 1:18 PM

Yup.

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#9
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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
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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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Re: Expansive Expansion: Newsletter Challenge (December 2013)

12/02/2013 9:58 AM

As doorman would say: yup.

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#8
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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
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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
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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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Doorman (4); James Stewart (1); prof peanut (1); RufusVS (2); SolarEagle (1); Tornado (2); trosenberger (1); Usbport (2); WilhelmHKoen (1)

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