Find the coefficient of x^7y^4 in the binomial expansion of (1/4x-16y)^11
Good heavens! I barely passed third year high school math -- and haven't taken a math class since then! If you notice, I only answer simple math questions.
Try searching for formulas online with the form (a+b)^n
Unfortunately, the website does not allow me to paste website links
To find the coefficient of a specific term in a binomial expansion, we can use the binomial theorem. The binomial theorem states that for any positive integer n, the expansion of (a + b)^n can be written as:
(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) * a^0 * b^n
where C(n, k) represents the binomial coefficient, which can be computed using the formula:
C(n, k) = n! / (k! * (n - k)!)
In the given expression, (1/4x - 16y)^11, the base is (1/4x - 16y), which means a = 1/4x and b = -16y. We are interested in finding the coefficient of x^7 * y^4, which corresponds to the term C(11, k) * (1/4x)^(11-k) * (-16y)^k, where k is the power of y in the term.
The power of x in the term is (11 - k), so we need to find a value of k that satisfies the equation 11 - k = 7. Solving this equation gives us k = 4.
Now, we can substitute the values of a, b, and k into the formula:
Coefficient = C(11, 4) * (1/4x)^(11-4) * (-16y)^4
Using the formula for binomial coefficient and simplifying the values of a and b, we have:
Coefficient = C(11, 4) * (1/4x)^7 * (-16y)^4
Coefficient = (11! / (4! * (11-4)!)) * (1/4)^7 * x^7 * (-16)^4 * y^4
Now, let's calculate the value of the coefficient:
Coefficient = (11! / (4! * 7!)) * (1/4)^7 * x^7 * (-16)^4 * y^4
After simplifying the factorials and exponentiations, we can compute the coefficient.