Use Pascal's triangle to expand the binomial. (d−5y)^6

Answer 1

To expand the binomial (d - 5y)^6 using Pascal's triangle, we can use the formula:

(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n

In this case, a = d and b = -5y. Therefore, the expanded form of (d - 5y)^6 is:

C(6, 0) * d^6 * (-5y)^0 + C(6, 1) * d^5 * (-5y)^1 + C(6, 2) * d^4 * (-5y)^2 + C(6, 3) * d^3 * (-5y)^3 + C(6, 4) * d^2 * (-5y)^4 + C(6, 5) * d^1 * (-5y)^5 + C(6, 6) * d^0 * (-5y)^6

To simplify this equation, we can use the values from Pascal's triangle:

C(6, 0) = 1, C(6, 1) = 6, C(6, 2) = 15, C(6, 3) = 20, C(6, 4) = 15, C(6, 5) = 6, C(6, 6) = 1

Replacing these values in the equation, we have:

1 * d^6 * (-5y)^0 + 6 * d^5 * (-5y)^1 + 15 * d^4 * (-5y)^2 + 20 * d^3 * (-5y)^3 + 15 * d^2 * (-5y)^4 + 6 * d^1 * (-5y)^5 + 1 * d^0 * (-5y)^6

Simplifying further, we have:

1 * d^6 * 1 + 6 * d^5 * (-5y) + 15 * d^4 * (-5y)^2 + 20 * d^3 * (-5y)^3 + 15 * d^2 * (-5y)^4 + 6 * d^1 * (-5y)^5 + 1 * (-5y)^6

Finally, we can simplify the binomial terms:

d^6 - 30d^5y + 300d^4y^2 - 1500d^3y^3 + 3750d^2y^4 - 4500dy^5 + 3125y^6

Therefore, the expanded form of (d - 5y)^6 using Pascal's triangle is:

d^6 - 30d^5y + 300d^4y^2 - 1500d^3y^3 + 3750d^2y^4 - 4500dy^5 + 3125y^6