If `n = (sqrt2 +1)^6`. Then the integer just greater than n is

Options

A. a) 199
B. b) 198
C. c) 197
D. d) 196

Answer

To solve the problem \( n = (\sqrt{2} + 1)^6 \) and find the integer just greater than \( n \), we will use the Binomial Theorem.

### Step-by-Step Solution:

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

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

3. **Combine All Terms**:
Now, we combine all the terms:
\[ n = 8 + 24\sqrt{2} + 60 + 40\sqrt{2} + 30 + 6\sqrt{2} + 1 \]
Combine the constant terms and the terms with \( \sqrt{2} \):
\[ n = (8 + 60 + 30 + 1) + (24\sqrt{2} + 40\sqrt{2} + 6\sqrt{2}) \]
\[ n = 99 + 70\sqrt{2} \]

4. **Substitute the Value of \( \sqrt{2} \)**:
Using \( \sqrt{2} \approx 1.414 \):
\[ n \approx 99 + 70 \cdot 1.414 \]
\[ n \approx 99 + 99.98 \approx 198.98 \]

5. **Find the Integer Just Greater than \( n \)**:
The integer just greater than \( n \) is:
\[ \lceil 198.98 \rceil = 199 \]

### Final Answer:
The integer just greater than \( n \) is **199**.