To determine the coefficient of [tex]\( x^3 y^2 \)[/tex] in the expansion of [tex]\( (2x + y)^5 \)[/tex], we will use the Binomial Theorem. The Binomial Theorem states:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
For the given expansion [tex]\((2x + y)^5\)[/tex], we identify:
- [tex]\( a = 2x \)[/tex]
- [tex]\( b = y \)[/tex]
- [tex]\( n = 5 \)[/tex]
We need to find the term in the expansion where the power of [tex]\( x \)[/tex] is 3 and the power of [tex]\( y \)[/tex] is 2. Let's denote this specific term as [tex]\( T_k \)[/tex].
In a general term [tex]\( T_k \)[/tex] in the expansion, the exponents should add up to [tex]\( n \)[/tex]. Therefore, we can write:
[tex]\[
T_k = \binom{5}{k} (2x)^{5-k} y^k
\][/tex]
We want the term where [tex]\( x^3 y^2 \)[/tex] appears:
- We need the power of [tex]\( x \)[/tex], which is [tex]\( (5 - k) \)[/tex], to be 3. Hence, [tex]\( 5 - k = 3 \)[/tex] which implies [tex]\( k = 2 \)[/tex].
- Therefore, the power of [tex]\( y \)[/tex], which is [tex]\( k \)[/tex], is 2.
Now substitute [tex]\( k = 2 \)[/tex] into the general term:
[tex]\[
T_2 = \binom{5}{2} (2x)^{5-2} y^2 = \binom{5}{2} (2x)^3 y^2
\][/tex]
Calculate the binomial coefficient:
[tex]\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
\][/tex]
Now simplify [tex]\((2x)^3\)[/tex]:
[tex]\[
(2x)^3 = 8x^3
\][/tex]
Hence, the term [tex]\( T_2 \)[/tex] becomes:
[tex]\[
T_2 = 10 \cdot 8x^3 y^2 = 80x^3 y^2
\][/tex]
Thus, the coefficient of [tex]\( x^3 y^2 \)[/tex] in the expansion of [tex]\( (2x + y)^5 \)[/tex] is:
[tex]\[
\boxed{80}
\][/tex]
