The greatest integer less than or equal to `(sqrt2+1)^6` is

Options

A. 197
B. 198
C. 196
D. 199

Explanation

To find the greatest integer less than or equal to \((\sqrt{2} + 1)^6\), we will use the Binomial Theorem to expand the expression and then find the integer part of the result.

### Step-by-step Solution:

1. **Use the Binomial Theorem**:
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
Here, we set \(a = \sqrt{2}\), \(b = 1\), and \(n = 6\). Thus, we can expand \((\sqrt{2} + 1)^6\):
\[ (\sqrt{2} + 1)^6 = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{2})^{6-k} (1)^k \]

2. **Calculate the individual terms**:
We will calculate each term in the expansion:
- For \(k = 0\): \(\binom{6}{0} (\sqrt{2})^6 = 1 \cdot 8 = 8\)
- For \(k = 1\): \(\binom{6}{1} (\sqrt{2})^5 = 6 \cdot 4\sqrt{2} = 24\sqrt{2}\)
- For \(k = 2\): \(\binom{6}{2} (\sqrt{2})^4 = 15 \cdot 4 = 60\)
- For \(k = 3\): \(\binom{6}{3} (\sqrt{2})^3 = 20 \cdot 2\sqrt{2} = 40\sqrt{2}\)
- For \(k = 4\): \(\binom{6}{4} (\sqrt{2})^2 = 15 \cdot 2 = 30\)
- For \(k = 5\): \(\binom{6}{5} (\sqrt{2})^1 = 6\sqrt{2}\)
- For \(k = 6\): \(\binom{6}{6} (1)^0 = 1\)

3. **Combine the terms**:
Now, we combine all these terms:
\[ (\sqrt{2} + 1)^6 = 8 + 24\sqrt{2} + 60 + 40\sqrt{2} + 30 + 6\sqrt{2} + 1 \]
Combine like terms:
\[ = (8 + 60 + 30 + 1) + (24\sqrt{2} + 40\sqrt{2} + 6\sqrt{2}) = 99 + 70\sqrt{2} \]

4. **Estimate \(\sqrt{2}\)**:
We know that \(\sqrt{2} \approx 1.414\). Thus, we can estimate:
\[ 70\sqrt{2} \approx 70 \cdot 1.414 \approx 99.98 \]
Therefore:
\[ 99 + 70\sqrt{2} \approx 99 + 99.98 \approx 198.98 \]

5. **Consider \((\sqrt{2} - 1)^6\)**:
We also note that \((\sqrt{2} - 1)\) is a small positive number (approximately \(0.414\)), and thus \((\sqrt{2} - 1)^6\) will be a very small positive number. This means:
\[ 0 < (\sqrt{2} - 1)^6 < 1 \]

6. **Final Calculation**:
Now, we can express:
\[ (\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 \approx 198.98 + \text{(a small positive number)} \]
Therefore, we can conclude:
\[ (\sqrt{2} + 1)^6 < 198.98 + 1 \]
Hence, the greatest integer less than or equal to \((\sqrt{2} + 1)^6\) is:
\[ \lfloor (\sqrt{2} + 1)^6 \rfloor = 197 \]

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