Find the coefficient of `x^5` in the expansion of `(1 + 2x)^6 (1-x)^7`.

Answer

To find the coefficient of \( x^5 \) in the expansion of \( (1 + 2x)^6 (1 - x)^7 \), we can follow these steps:

### Step 1: Expand \( (1 + 2x)^6 \)

Using the Binomial Theorem, we can expand \( (1 + 2x)^6 \):

\[ (1 + 2x)^6 = \sum_{k=0}^{6} \binom{6}{k} (2x)^k = \sum_{k=0}^{6} \binom{6}{k} 2^k x^k \]

This gives us:

\[ = \binom{6}{0} + \binom{6}{1} \cdot 2x + \binom{6}{2} \cdot (2x)^2 + \binom{6}{3} \cdot (2x)^3 + \binom{6}{4} \cdot (2x)^4 + \binom{6}{5} \cdot (2x)^5 + \binom{6}{6} \cdot (2x)^6 \]

Calculating the coefficients:

- \( \binom{6}{0} = 1 \)
- \( \binom{6}{1} = 6 \)
- \( \binom{6}{2} = 15 \)
- \( \binom{6}{3} = 20 \)
- \( \binom{6}{4} = 15 \)
- \( \binom{6}{5} = 6 \)
- \( \binom{6}{6} = 1 \)

Thus, we have:

\[ (1 + 2x)^6 = 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^6 \]

### Step 2: Expand \( (1 - x)^7 \)

Using the Binomial Theorem again, we expand \( (1 - x)^7 \):

\[ (1 - x)^7 = \sum_{j=0}^{7} \binom{7}{j} (-x)^j = \sum_{j=0}^{7} \binom{7}{j} (-1)^j x^j \]

This gives us:

\[ = \binom{7}{0} - \binom{7}{1} x + \binom{7}{2} x^2 - \binom{7}{3} x^3 + \binom{7}{4} x^4 - \binom{7}{5} x^5 + \binom{7}{6} x^6 - \binom{7}{7} x^7 \]

Calculating the coefficients:

- \( \binom{7}{0} = 1 \)
- \( \binom{7}{1} = 7 \)
- \( \binom{7}{2} = 21 \)
- \( \binom{7}{3} = 35 \)
- \( \binom{7}{4} = 35 \)
- \( \binom{7}{5} = 21 \)
- \( \binom{7}{6} = 7 \)
- \( \binom{7}{7} = 1 \)

Thus, we have:

\[ (1 - x)^7 = 1 - 7x + 21x^2 - 35x^3 + 35x^4 - 21x^5 + 7x^6 - x^7 \]

### Step 3: Combine the Expansions

Now we need to find the coefficient of \( x^5 \) in the product \( (1 + 2x)^6 (1 - x)^7 \).

To find the coefficient of \( x^5 \), we consider the combinations of terms from both expansions that yield \( x^5 \):

1. Constant term from \( (1 + 2x)^6 \) and \( x^5 \) from \( (1 - x)^7 \):
\[ 1 \cdot (-21) = -21 \]

2. Coefficient of \( x \) from \( (1 + 2x)^6 \) and \( x^4 \) from \( (1 - x)^7 \):
\[ 12 \cdot 35 = 420 \]

3. Coefficient of \( x^2 \) from \( (1 + 2x)^6 \) and \( x^3 \) from \( (1 - x)^7 \):
\[ 60 \cdot (-35) = -2100 \]

4. Coefficient of \( x^3 \) from \( (1 + 2x)^6 \) and \( x^2 \) from \( (1 - x)^7 \):
\[ 160 \cdot 21 = 3360 \]

5. Coefficient of \( x^4 \) from \( (1 + 2x)^6 \) and \( x \) from \( (1 - x)^7 \):
\[ 240 \cdot (-7) = -1680 \]

6. Coefficient of \( x^5 \) from \( (1 + 2x)^6 \) and constant term from \( (1 - x)^7 \):
\[ 192 \cdot 1 = 192 \]

### Step 4: Sum the Contributions

Now we sum all these contributions to find the coefficient of \( x^5 \):

\[ -21 + 420 - 2100 + 3360 - 1680 + 192 \]

Calculating this step-by-step:

1. \( -21 + 420 = 399 \)
2. \( 399 - 2100 = -1701 \)
3. \( -1701 + 3360 = 1659 \)
4. \( 1659 - 1680 = -21 \)
5. \( -21 + 192 = 171 \)

Thus, the coefficient of \( x^5 \) in the expansion of \( (1 + 2x)^6 (1 - x)^7 \) is \( 171 \).

### Final Answer:
The coefficient of \( x^5 \) is \( 171 \).

---