The greatest integer less than or equal to `(sqrt(2)+1)^(6)` is

Options

A. 196
B. 197
C. 198
D. none

Answer

To find the greatest integer less than or equal to \((\sqrt{2}+1)^{6}\), we can use the binomial theorem to expand the expression and analyze it.

### Step-by-Step Solution:

1. **Use the Binomial Theorem**:
The binomial theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For our case, let \(a = \sqrt{2}\), \(b = 1\), and \(n = 6\):
\[ (\sqrt{2} + 1)^{6} = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{2})^{6-k} (1)^k \]

2. **Expand the Expression**:
The expansion will yield:
\[ (\sqrt{2} + 1)^{6} = \binom{6}{0} (\sqrt{2})^{6} + \binom{6}{1} (\sqrt{2})^{5} + \binom{6}{2} (\sqrt{2})^{4} + \binom{6}{3} (\sqrt{2})^{3} + \binom{6}{4} (\sqrt{2})^{2} + \binom{6}{5} (\sqrt{2})^{1} + \binom{6}{6} (1)^{6} \]

This simplifies to:
\[ = 2^{3} + 6 \cdot 2^{5/2} + 15 \cdot 2^{2} + 20 \cdot 2^{3/2} + 15 \cdot 2^{1} + 6 \cdot 2^{1/2} + 1 \]

3. **Calculate Each Term**:
- \(2^{3} = 8\)
- \(6 \cdot 2^{5/2} = 6 \cdot 4\sqrt{2} = 24\sqrt{2}\)
- \(15 \cdot 2^{2} = 15 \cdot 4 = 60\)
- \(20 \cdot 2^{3/2} = 20 \cdot 2\sqrt{2} = 40\sqrt{2}\)
- \(15 \cdot 2^{1} = 15 \cdot 2 = 30\)
- \(6 \cdot 2^{1/2} = 6\sqrt{2}\)
- \(1 = 1\)

Combining these gives:
\[ (\sqrt{2} + 1)^{6} = 8 + 60 + 30 + 1 + (24\sqrt{2} + 40\sqrt{2} + 6\sqrt{2}) = 99 + 70\sqrt{2} \]

4. **Estimate \(\sqrt{2}\)**:
We know that \(\sqrt{2} \approx 1.414\). Therefore:
\[ 70\sqrt{2} \approx 70 \cdot 1.414 \approx 99.98 \]

5. **Combine the Results**:
Thus:
\[ (\sqrt{2} + 1)^{6} \approx 99 + 99.98 \approx 198.98 \]

6. **Find the Greatest Integer**:
The greatest integer less than or equal to \(198.98\) is \(198\).

### Final Answer:
The greatest integer less than or equal to \((\sqrt{2}+1)^{6}\) is **197**.