The nubmber of non - zeroes terns in the expansion of `(1+sqrt(5))^(6)+(sqrt(5)-1)^(6)` is

Options

A. 3
B. 4
C. 5
D. 0

Answer

To find the number of non-zero terms in the expression \((1+\sqrt{5})^6 + (\sqrt{5}-1)^6\), we can use the Binomial Theorem to expand both terms.

### Step-by-Step Solution:

1. **Apply the Binomial Theorem**:
The Binomial Theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). We will apply this to both \((1+\sqrt{5})^6\) and \((\sqrt{5}-1)^6\).

2. **Expand \((1+\sqrt{5})^6\)**:
\[ (1+\sqrt{5})^6 = \sum_{k=0}^{6} \binom{6}{k} 1^{6-k} (\sqrt{5})^k = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{5})^k \]
This gives us the terms:
- For \(k=0\): \(\binom{6}{0} (\sqrt{5})^0 = 1\)
- For \(k=1\): \(\binom{6}{1} (\sqrt{5})^1 = 6\sqrt{5}\)
- For \(k=2\): \(\binom{6}{2} (\sqrt{5})^2 = 15 \cdot 5 = 75\)
- For \(k=3\): \(\binom{6}{3} (\sqrt{5})^3 = 20 \cdot 5\sqrt{5} = 100\sqrt{5}\)
- For \(k=4\): \(\binom{6}{4} (\sqrt{5})^4 = 15 \cdot 25 = 375\)
- For \(k=5\): \(\binom{6}{5} (\sqrt{5})^5 = 6 \cdot 5\sqrt{5} = 30\sqrt{5}\)
- For \(k=6\): \(\binom{6}{6} (\sqrt{5})^6 = 1 \cdot 125 = 125\)

3. **Expand \((\sqrt{5}-1)^6\)**:
\[ (\sqrt{5}-1)^6 = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{5})^k (-1)^{6-k} \]
This gives us the terms:
- For \(k=0\): \(\binom{6}{0} (\sqrt{5})^0 (-1)^6 = 1\)
- For \(k=1\): \(\binom{6}{1} (\sqrt{5})^1 (-1)^5 = -6\sqrt{5}\)
- For \(k=2\): \(\binom{6}{2} (\sqrt{5})^2 (-1)^4 = 15 \cdot 5 = 75\)
- For \(k=3\): \(\binom{6}{3} (\sqrt{5})^3 (-1)^3 = -20 \cdot 5\sqrt{5} = -100\sqrt{5}\)
- For \(k=4\): \(\binom{6}{4} (\sqrt{5})^4 (-1)^2 = 15 \cdot 25 = 375\)
- For \(k=5\): \(\binom{6}{5} (\sqrt{5})^5 (-1)^1 = -6 \cdot 5\sqrt{5} = -30\sqrt{5}\)
- For \(k=6\): \(\binom{6}{6} (\sqrt{5})^6 (-1)^0 = 1 \cdot 125 = 125\)

4. **Combine the Two Expansions**:
Now we add the two expansions:
\[ (1+\sqrt{5})^6 + (\sqrt{5}-1)^6 \]
- Constant term: \(1 + 1 = 2\)
- Coefficient of \(\sqrt{5}\): \(6\sqrt{5} - 6\sqrt{5} = 0\)
- Coefficient of \((\sqrt{5})^2\): \(75 + 75 = 150\)
- Coefficient of \((\sqrt{5})^3\): \(100\sqrt{5} - 100\sqrt{5} = 0\)
- Coefficient of \((\sqrt{5})^4\): \(375 + 375 = 750\)
- Coefficient of \((\sqrt{5})^5\): \(30\sqrt{5} - 30\sqrt{5} = 0\)
- Coefficient of \((\sqrt{5})^6\): \(125 + 125 = 250\)

5. **Identify Non-Zero Terms**:
The resulting expression is:
\[ 2 + 150 + 750 + 250 \]
The non-zero terms correspond to the powers of \(\sqrt{5}\) that are even: \(0, 2, 4, 6\).

6. **Count Non-Zero Terms**:
The non-zero terms are for \(k=0, 2, 4, 6\), which gives us a total of **4 non-zero terms**.

### Final Answer:
The number of non-zero terms in the expansion of \((1+\sqrt{5})^6 + (\sqrt{5}-1)^6\) is **4**.