Expand `(2x+y)^(5)` with the help of binomial theorem

Explanation

To expand \((2x + y)^5\) using the Binomial Theorem, we follow these steps:

### Step 1: Identify the terms in the binomial expression
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, \(a = 2x\), \(b = y\), and \(n = 5\).

### Step 2: Write the expansion using the Binomial Theorem
Using the theorem, we can write:
\[ (2x + y)^5 = \sum_{k=0}^{5} \binom{5}{k} (2x)^{5-k} y^k \]

### Step 3: Calculate each term in the expansion
We will calculate each term for \(k = 0\) to \(k = 5\):

1. **For \(k = 0\)**:
\[ \binom{5}{0} (2x)^5 y^0 = 1 \cdot (2x)^5 \cdot 1 = 32x^5 \]

2. **For \(k = 1\)**:
\[ \binom{5}{1} (2x)^4 y^1 = 5 \cdot (16x^4) \cdot y = 80x^4y \]

3. **For \(k = 2\)**:
\[ \binom{5}{2} (2x)^3 y^2 = 10 \cdot (8x^3) \cdot y^2 = 80x^3y^2 \]

4. **For \(k = 3\)**:
\[ \binom{5}{3} (2x)^2 y^3 = 10 \cdot (4x^2) \cdot y^3 = 40x^2y^3 \]

5. **For \(k = 4\)**:
\[ \binom{5}{4} (2x)^1 y^4 = 5 \cdot (2x) \cdot y^4 = 10xy^4 \]

6. **For \(k = 5\)**:
\[ \binom{5}{5} (2x)^0 y^5 = 1 \cdot 1 \cdot y^5 = y^5 \]

### Step 4: Combine all the terms
Now, we combine all the calculated terms:
\[ (2x + y)^5 = 32x^5 + 80x^4y + 80x^3y^2 + 40x^2y^3 + 10xy^4 + y^5 \]

### Final Answer
Thus, the expansion of \((2x + y)^5\) is:
\[ 32x^5 + 80x^4y + 80x^3y^2 + 40x^2y^3 + 10xy^4 + y^5 \]