the tree Posted September 13, 2005 Posted September 13, 2005 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?
matt grime Posted September 13, 2005 Posted September 13, 2005 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.
insane_alien Posted September 13, 2005 Posted September 13, 2005 i'm sure you'll get onto that next in maths. thats what happened in my class. and don't worry you'll only get to about the seventh layer for a pascals triangle
DQW Posted September 13, 2005 Posted September 13, 2005 deleted... ...I think I may have posted exactly what matt had left as an exercise for the OP.
Mobius Posted September 14, 2005 Posted September 14, 2005 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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