Evalute: `(2+sqrt(3))^(7)+(2-sqrt(3))^(7)`

Answer

To evaluate the expression \((2+\sqrt{3})^{7} + (2-\sqrt{3})^{7}\), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]

### Step-by-Step Solution:

1. **Apply the Binomial Theorem**:
We will first expand \((2+\sqrt{3})^{7}\) and \((2-\sqrt{3})^{7}\) using the Binomial Theorem.

\[ (2+\sqrt{3})^{7} = \sum_{k=0}^{7} \binom{7}{k} 2^{7-k} (\sqrt{3})^k \]
\[ (2-\sqrt{3})^{7} = \sum_{k=0}^{7} \binom{7}{k} 2^{7-k} (-\sqrt{3})^k \]

2. **Combine the Two Expansions**:
Now we add these two expansions together:

\[ (2+\sqrt{3})^{7} + (2-\sqrt{3})^{7} = \sum_{k=0}^{7} \binom{7}{k} 2^{7-k} (\sqrt{3})^k + \sum_{k=0}^{7} \binom{7}{k} 2^{7-k} (-\sqrt{3})^k \]

Notice that when \(k\) is odd, the terms will cancel out because one will be positive and the other negative. When \(k\) is even, the terms will add up.

3. **Identify Even Terms**:
We will only consider the even \(k\) terms from the expansion:

\[ = \sum_{k \text{ even}} \binom{7}{k} 2^{7-k} (\sqrt{3})^k + \sum_{k \text{ even}} \binom{7}{k} 2^{7-k} (-\sqrt{3})^k \]

This simplifies to:

\[ = 2 \sum_{k \text{ even}} \binom{7}{k} 2^{7-k} (\sqrt{3})^k \]

4. **Calculate the Even Terms**:
The even values of \(k\) are \(0, 2, 4, 6\). We will calculate these terms:

- For \(k=0\):
\[ \binom{7}{0} 2^{7} (\sqrt{3})^{0} = 1 \cdot 128 \cdot 1 = 128 \]
- For \(k=2\):
\[ \binom{7}{2} 2^{5} (\sqrt{3})^{2} = 21 \cdot 32 \cdot 3 = 2016 \]
- For \(k=4\):
\[ \binom{7}{4} 2^{3} (\sqrt{3})^{4} = 35 \cdot 8 \cdot 9 = 2520 \]
- For \(k=6\):
\[ \binom{7}{6} 2^{1} (\sqrt{3})^{6} = 7 \cdot 2 \cdot 27 = 378 \]

5. **Sum the Even Terms**:
Now we sum these contributions:

\[ 128 + 2016 + 2520 + 378 = 5042 \]

6. **Final Calculation**:
Since we have \(2\) times the sum of the even terms:

\[ (2+\sqrt{3})^{7} + (2-\sqrt{3})^{7} = 2 \cdot 5042 = 10084 \]

### Final Answer:
\[ (2+\sqrt{3})^{7} + (2-\sqrt{3})^{7} = 10084 \]