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

Options

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

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.

### Step-by-step Solution:

1. **Identify the expression**: We need to evaluate \((\sqrt{2} + 1)^6\).

2. **Use the Binomial Theorem**: According to the binomial theorem, \((a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\). Here, \(a = \sqrt{2}\), \(b = 1\), and \(n = 6\).

3. **Expand the expression**:
\[ (\sqrt{2} + 1)^6 = \sum_{r=0}^{6} \binom{6}{r} (\sqrt{2})^{6-r} (1)^r \]
This simplifies to:
\[ = \binom{6}{0} (\sqrt{2})^6 + \binom{6}{1} (\sqrt{2})^5 + \binom{6}{2} (\sqrt{2})^4 + \binom{6}{3} (\sqrt{2})^3 + \binom{6}{4} (\sqrt{2})^2 + \binom{6}{5} (\sqrt{2})^1 + \binom{6}{6} (\sqrt{2})^0 \]

4. **Calculate each term**:
- \(\binom{6}{0} (\sqrt{2})^6 = 1 \cdot 8 = 8\)
- \(\binom{6}{1} (\sqrt{2})^5 = 6 \cdot 4\sqrt{2} = 24\sqrt{2}\)
- \(\binom{6}{2} (\sqrt{2})^4 = 15 \cdot 4 = 60\)
- \(\binom{6}{3} (\sqrt{2})^3 = 20 \cdot 2\sqrt{2} = 40\sqrt{2}\)
- \(\binom{6}{4} (\sqrt{2})^2 = 15 \cdot 2 = 30\)
- \(\binom{6}{5} (\sqrt{2})^1 = 6\sqrt{2}\)
- \(\binom{6}{6} (\sqrt{2})^0 = 1\)

5. **Combine the terms**:
\[ (\sqrt{2} + 1)^6 = 8 + 60 + 30 + 1 + (24\sqrt{2} + 40\sqrt{2} + 6\sqrt{2}) \]
\[ = 99 + 70\sqrt{2} \]

6. **Approximate \(\sqrt{2}\)**: We know that \(\sqrt{2} \approx 1.414\). Thus,
\[ 70\sqrt{2} \approx 70 \cdot 1.414 = 99.98 \]

7. **Calculate the total**:
\[ 99 + 99.98 \approx 198.98 \]

8. **Find the greatest integer less than or equal to this value**:
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 \(\boxed{198}\).