Question
Find the binomial expansion when (2x+3y) to the power 9.
Answers
GPT 3.5
The binomial expansion of (2x+3y)^9 can be found using the binomial theorem.
The binomial theorem states that for any positive integer n, the expansion of (a+b)^n is given by:
(a+b)^n = nC0 * a^(n-0) * b^0 + nC1 * a^(n-1) * b^1 + nC2 * a^(n-2) * b^2 + ... + nC(n-1) * a^1 * b^(n-1) + nCn * a^0 * b^n
Where nCr is the binomial coefficient, given by n! / (r! * (n-r)!)
In this case, a = 2x and b = 3y, so the expression becomes:
(2x+3y)^9 = 9C0 * (2x)^9 * (3y)^0 + 9C1 * (2x)^8 * (3y)^1 + 9C2 * (2x)^7 * (3y)^2 + ... + 9C7 * (2x)^2 * (3y)^7 + 9C8 * (2x)^1 * (3y)^8 + 9C9 * (2x)^0 * (3y)^9
Simplifying the coefficients and exponents, the expansion becomes:
(2x+3y)^9 = 1 * 2^9 * x^9 + 9 * 2^8 * 3^1 * x^8 * y^1 + 36 * 2^7 * 3^2 * x^7 * y^2 + ... + 84 * 2^2 * 3^7 * x^2 * y^7 + 72 * 2^1 * 3^8 * x^1 * y^8 + 27 * 3^9 * y^9
Simplifying further, we have:
(2x+3y)^9 = 512x^9 + 4608x^8y + 19440x^7y^2 + ... + 52488x^2y^7 + 23328xy^8 + 19683y^9
The binomial theorem states that for any positive integer n, the expansion of (a+b)^n is given by:
(a+b)^n = nC0 * a^(n-0) * b^0 + nC1 * a^(n-1) * b^1 + nC2 * a^(n-2) * b^2 + ... + nC(n-1) * a^1 * b^(n-1) + nCn * a^0 * b^n
Where nCr is the binomial coefficient, given by n! / (r! * (n-r)!)
In this case, a = 2x and b = 3y, so the expression becomes:
(2x+3y)^9 = 9C0 * (2x)^9 * (3y)^0 + 9C1 * (2x)^8 * (3y)^1 + 9C2 * (2x)^7 * (3y)^2 + ... + 9C7 * (2x)^2 * (3y)^7 + 9C8 * (2x)^1 * (3y)^8 + 9C9 * (2x)^0 * (3y)^9
Simplifying the coefficients and exponents, the expansion becomes:
(2x+3y)^9 = 1 * 2^9 * x^9 + 9 * 2^8 * 3^1 * x^8 * y^1 + 36 * 2^7 * 3^2 * x^7 * y^2 + ... + 84 * 2^2 * 3^7 * x^2 * y^7 + 72 * 2^1 * 3^8 * x^1 * y^8 + 27 * 3^9 * y^9
Simplifying further, we have:
(2x+3y)^9 = 512x^9 + 4608x^8y + 19440x^7y^2 + ... + 52488x^2y^7 + 23328xy^8 + 19683y^9