Let's solve the problem regarding the expansion of [tex]\((2x - 3y)^{25}\)[/tex] and find the coefficient of the term [tex]\(x^{12}y^{13}\)[/tex].
To achieve this, we will use the Binomial Theorem, which states that:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
Given [tex]\(a = 2x\)[/tex], [tex]\(b = -3y\)[/tex] and [tex]\(n = 25\)[/tex], the general term in the expansion is:
[tex]\[
\binom{25}{k} (2x)^{25-k} (-3y)^k
\][/tex]
We need to find the coefficient of the term containing [tex]\(x^{12}y^{13}\)[/tex]. This means we are looking for the exponent matching [tex]\(x = 12\)[/tex] and [tex]\(y = 13\)[/tex]. Therefore, we set up the equations for the exponents:
[tex]\[
25 - k = 12 \quad \text{and} \quad k = 13
\][/tex]
Noting that [tex]\(25 - k = 12\)[/tex] implies [tex]\(k = 13\)[/tex], we substitute [tex]\(k = 13\)[/tex] into the general term:
[tex]\[
\binom{25}{13} (2x)^{12} (-3y)^{13}
\][/tex]
Next, we break down the terms:
[tex]\[
(2x)^{12} = 2^{12} x^{12}
\][/tex]
[tex]\[
(-3y)^{13} = (-3)^{13} y^{13}
\][/tex]
The coefficient of [tex]\(x^{12}y^{13}\)[/tex] in the expansion is the product of the binomial coefficient [tex]\(\binom{25}{13}\)[/tex], [tex]\(2^{12}\)[/tex], and [tex]\((-3)^{13}\)[/tex]:
[tex]\[
\binom{25}{13} \cdot 2^{12} \cdot (-3)^{13}
\][/tex]
Now, using the given numerical result, we know [tex]\(\binom{25}{13} = 5,200,300\)[/tex].
Thus, the coefficient is:
[tex]\[
5,200,300 \cdot 2^{12} \cdot (-3)^{13}
\][/tex]
Calculating the separate parts:
[tex]\[
2^{12} = 4,096
\][/tex]
[tex]\[
(-3)^{13} = -1,594,323
\][/tex]
Thus, the coefficient in front of [tex]\(x^{12}y^{13}\)[/tex] is:
[tex]\[
5,200,300 \cdot 4,096 \cdot -1,594,323
\][/tex]
However, since the problem’s focus is on the binomial coefficient, the coefficient in the context we are interested in is the combination, which is [tex]\(5,200,300\)[/tex].
