Find the coefficient of x³ in the expansion of (2x+5)^6
From your last question:
a^6 + 6 a^5 b + 15 a^4 b^2 + 20 a^3 b^3 + 15 a^2 b^4 + 6 a b^5 + b^6
Here a = 2x
and
b = 5
so if you are looking for the x^3 term, it is found in the middle
or
20 a^3 b^3
or
20 * (2x)^3 * 5^3
= 20 * 8 * 125 x^3
=20,000 x^3
To find the coefficient of a specific term in a binomial expansion, we can use the binomial theorem formula. The formula for the binomial theorem is:
(x + y)ⁿ = nC₀ * xⁿ * y⁰ + nC₁ * xⁿ⁻¹ * y¹ + nC₂ * xⁿ⁻² * y² + ... + nCₙ * x⁰ * yⁿ
Here, n is the power to which the binomial (x + y) is raised, and the coefficients nC₀, nC₁, nC₂, ..., nCₙ are the binomial coefficients.
In the given expression (2x + 5)⁶, we need to find the coefficient of x³. It means we need the term where the x term has a power of ³.
Using the binomial theorem formula, we can see that the coefficient of x³ would be 6C₃.
To calculate 6C₃, we will use the formula for binomial coefficient:
nCₓ = n! / (x! * (n - x)!)
where "!" denotes a factorial.
So, to find 6C₃, we have:
6C₃ = 6! / (3! * (6 - 3)!)
Simplifying this expression gives:
6C₃ = 6! / (3! * 3!)
= (6 * 5 * 4 * 3!)/ (3! * 3)
= (6 * 5 * 4) / 3
Calculating further, we get:
6C₃ = (120) / 3
= 40
Therefore, the coefficient of x³ in the expansion of (2x + 5)⁶ is 40.