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nth layer of pascals triangle.


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Today my maths teacher showed us how we can expand [math](x+1)^n[/math] using pascals triangle.

 

For instance, to find [math](x+1)^4[/math], we know that the fourth layer of pascals triangle is 1,4,6,4,1 and these numbers come into the awnser:[math]x^{4}+4x^{3}+6x^{2}+4x+1[/math], the other bits to be put in are fairly simple.

 

This was pretty impressive but if the exponent were 15 I wouldn't really want to draw up a 15 layered pascal triangle, I'm to lazy.

 

So my question is: is there a way I can work out the [math]n^{th}[/math] layer of Pascals triangle without working down the whole thing?

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yes, the entries are called binomial coefficients.

 

nCr is the r'th number in the n'th row as r goes from 0 to n

 

i think you should figure out the formual yourself by thinking how when you expand

 

(1+x)(1+x)(1+x)...(1+x)

 

 

you can obtain x^r: hint it's lik picking the x from r brackets and the 1 from the remaining brackets.

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Pascal's trinagle is not just good at finding coeficients of polynomials.

If you add up each row the numbers increase by powers of two...

If you split the rows into diagonals and add them they give you the Fibonacci sequence.

There are loads more examples where the triangle shows the relationship between numbers and elementary number patterns.

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