how would you set this problem up using pascal's triangle?

(x+y)5 (raised to the 5th power)

5th row, 6 terms.

http://en.wikipedia.org/wiki/Pascal's_triangle

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

so (x+y)^5
= x^5 + 5x^4y + 10x^3y^2 + .... + y^5

To set up the problem using Pascal's triangle, you can expand the expression (x+y)^5 by using the coefficients from the 5th row of Pascal's triangle.

The 5th row of Pascal's triangle is: 1 5 10 10 5 1

To expand (x+y)^5, you will have 6 terms because there are 6 elements in the 5th row of Pascal's triangle.

The general formula for expanding (x+y)^n using Pascal's triangle is:
(x+y)^n = C(n,0)x^n + C(n,1)x^(n-1)y + C(n,2)x^(n-2)y^2 + ... + C(n,n-1)xy^(n-1) + C(n,n)y^n

Using the 5th row of Pascal's triangle, you can set up the expansion for (x+y)^5 as follows:

(x+y)^5 = C(5,0)x^5 + C(5,1)x^4y + C(5,2)x^3y^2 + C(5,3)x^2y^3 + C(5,4)xy^4 + C(5,5)y^5

Now, you can substitute the values from Pascal's triangle for the binomial coefficients:

(x+y)^5 = 1 * x^5 + 5 * x^4y + 10 * x^3y^2 + 10 * x^2y^3 + 5 * xy^4 + 1 * y^5

Simplifying this expression will give you the expanded form of (x+y)^5 using Pascal's triangle.