Write out the expansions of the following:
(e ) `(1+2x)^(7)`

Answer

To expand the expression \( (1 + 2x)^7 \) using the Binomial Theorem, we can follow these steps:

### Step 1: Identify the components of the binomial expansion
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, \( a = 1 \), \( b = 2x \), and \( n = 7 \).

### Step 2: Write out the expansion
Using the Binomial Theorem, we can write the expansion as:
\[ (1 + 2x)^7 = \sum_{k=0}^{7} \binom{7}{k} (1)^{7-k} (2x)^k \]
This simplifies to:
\[ = \sum_{k=0}^{7} \binom{7}{k} (2x)^k \]

### Step 3: Calculate each term in the expansion
Now, we will calculate each term for \( k = 0 \) to \( k = 7 \):

1. For \( k = 0 \):
\[ \binom{7}{0} (2x)^0 = 1 \cdot 1 = 1 \]

2. For \( k = 1 \):
\[ \binom{7}{1} (2x)^1 = 7 \cdot 2x = 14x \]

3. For \( k = 2 \):
\[ \binom{7}{2} (2x)^2 = 21 \cdot (2^2 x^2) = 21 \cdot 4x^2 = 84x^2 \]

4. For \( k = 3 \):
\[ \binom{7}{3} (2x)^3 = 35 \cdot (2^3 x^3) = 35 \cdot 8x^3 = 280x^3 \]

5. For \( k = 4 \):
\[ \binom{7}{4} (2x)^4 = 35 \cdot (2^4 x^4) = 35 \cdot 16x^4 = 560x^4 \]

6. For \( k = 5 \):
\[ \binom{7}{5} (2x)^5 = 21 \cdot (2^5 x^5) = 21 \cdot 32x^5 = 672x^5 \]

7. For \( k = 6 \):
\[ \binom{7}{6} (2x)^6 = 7 \cdot (2^6 x^6) = 7 \cdot 64x^6 = 448x^6 \]

8. For \( k = 7 \):
\[ \binom{7}{7} (2x)^7 = 1 \cdot (2^7 x^7) = 1 \cdot 128x^7 = 128x^7 \]

### Step 4: Combine all the terms
Now, we can combine all the terms we calculated:
\[ (1 + 2x)^7 = 1 + 14x + 84x^2 + 280x^3 + 560x^4 + 672x^5 + 448x^6 + 128x^7 \]

### Final Answer
The expansion of \( (1 + 2x)^7 \) is:
\[ 1 + 14x + 84x^2 + 280x^3 + 560x^4 + 672x^5 + 448x^6 + 128x^7 \]