Why 11 and binomial theorem ?

[Sriman Dutta]

I observe a relationship here.....

Binomial theorem depends on Pascal's triangle which goes as follows:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

 

But, (11)^n also shows a similar sequence.

 

Is there any relationship between these two ?

[fiveworlds]

Okay but after n=4 it fails completely ...

[Sriman Dutta]

Hmmm.....Right.

[Endy0816]

Magic 11's is what you've happened upon.

 

1 5 10 10 5 1 -> 1 + 50 + 10*100 + 10*1000 + 5*10000 + 1*100000 = 161051

 

http://jwilson.coe.uga.edu/EMAT6680Su12/Berryman/6690/BerrymanK-Pascals/BerrymanK-Pascals.html

 

 

[Country Boy]

Basically, this is because 11= 10+ 1. [math]11^2= (10+ 1)^2= 1(10^2)+ 2(10)(1)+ 1(1)= 100+ 20+ 1= 121[/math]. [math]11^3= (10+ 1)^3= 10^3+ 3(10^2)(1)+ 3(10)(1^2)+ 1^3= 100+ 300+ 30= 1= 1331[/math]. [math]11^4= 1(10^4)+ 4(10^3)(1)+ 6(10^2)(1^2)+ 6(10)(1^3)+ 1(1^4)= 10000+ 4000+ 600+ 40+ 1= 14641[/math]. it fails for n> 4 because the binomial coefficients then have more than 1 digit.

[DrKrettin]

That calculation looks as if it might be independent of the numbering system, so would be valid in, say, hexadecimal. If so, I guess it would work there for n=5 and until a coefficient had more than 1 digit in hex.

[Sriman Dutta]

Magic 11's is what you've happened upon.

 

1 5 10 10 5 1 -> 1 + 50 + 10*100 + 10*1000 + 5*10000 + 1*100000 = 161051

 

http://jwilson.coe.uga.edu/EMAT6680Su12/Berryman/6690/BerrymanK-Pascals/BerrymanK-Pascals.html

 

 

Thanks Endy. It's really very helpful. I didn't knew that the Pascal's Triangle has so many interesting things.

[Endy0816]

Thanks Endy. It's really very helpful. I didn't knew that the Pascal's Triangle has so many interesting things.

 

Welcome.

 

Reminded me of a similar pattern seen in Lychrel numbers.

 

196 + 691 = 7 18 7 (887)

887 + 788 = 15 16 15 (1675)

1675 + 5761 = 6 13 13 6 (7436)