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About LinkBacksWhat are 2 examples with detailed solutions of the application of mathematics for Pascal's triangle/ theorem???
like i said last night, i don't know any applications for Pascal's theorem, but i can give you two examples for pascal's triangleOriginally Posted by Sarah_Arthur
pascal's triangle is particularly useful in expanding binomials, for example,
we know that
(x + 3)^4 = x^4 + 12x^3 + 54x^2 + 108x + 81
and
(x + 2)^6 = x^6 + 12x^5 + 60x^4 + 160x^3 + 240x^2 + 192x + 64
using pascal's triangle to find the coefficients of all the terms
do you know how to do this?
the diagram here PascalTriangle is known as pascal's triangle.
this is how you construct it. start with a triangle of three 1's as you see at the top
for every line you begin and end with a 1. for the numbers between, you get them by adding the two numbers above them
example, look at the line 1 2 1. i start with a 1 and end with a 1, for the middle, i add the two 1's above and get 2
for the line 1 4 6 4 1
i start with a one, then add the 3 and 1 above to get the 4 then i add the 3 and 3 above the next space to get the 6 then i add the 3 and 1 above the next space to get 4 and then i end with a 1 again.
these lines tell you the coefficient of binomial expansions.
this is why (x + 1)^2 = 1x^2 + 2x + 1 ....because of the line 1 2 1
let's see how this works.
if we have some binomial say (x + 3)^3 , i look for the line where 3 is the second number. that means the line 1 3 3 1
so these are going to be the coefficients of the terms in the expansion. as you see there will be 4 terms.
but what about the x and the 3, this is how you deal with them.
you take the first term is the binomial (by the way, a binomial just means the sum of 2 terms) which in this case is an x, and you count down with the powers in each term, starting with the power you are trying to expand by. so for (x + 3)^3, we are going to have:
x^3 in the first term
x^2 in the second term
x^1 in the third term
x^0 in the fourth term (note that x^0 = 1, that's why you don't see x in the last term)
now for the second number, in this case the 3, you count up with the powers starting at zero going up to the power you're expanding by. so we will get:
3^0 = 1 in the first term
3^1 = 3 in the second term
3^2 = 9 in the thrd term
3^3 = 27 in the fourth term
for each term we will multiply the corresponding numbers
so we have
(x^3)(3^0) for the first term
(x^2)(3^1) for the second term
(x)(3^2) for the third term
(x^0)(3^3) for the fourth term
so you see we count down with the powers of the first number, and up with the powers of the second number.
and then we use pascal's triangle for the coefficients
so we have:
1(x^3)(3^0) for the first term
3(x^2)(3^1) for the second term
3(x)(3^2) for the third term
1(x^0)(3^3) for the fourth term
we get 1 3 3 1 from pascal's triangle. now we just add them together, so:
(x + 3)^3 = 1(x^3)(3^0) + 3(x^2)(3^1) + 3(x)(3^2) + 1(x^0)(3^3)
= 1x^3(1) + 3(3)x^2 + 3(9)x + 1(1)(27)
= x^3 + 9x^2 + 27x + 27
got it?
see http://en.wikipedia.org/wiki/Pascal's_triangle for more info if you want, but i think you have everything you need to know above
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