If [ ] denotes the greatest integer function, then `[( sqrt2 + 1)^(6))` is equal

Options

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

Answer

To solve the problem of finding \(\left\lfloor (\sqrt{2} + 1)^6 \right\rfloor\), we will follow these steps:

### Step 1: Expand \((\sqrt{2} + 1)^6\) using the Binomial Theorem
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, \(a = \sqrt{2}\), \(b = 1\), and \(n = 6\).

Thus, we can write:
\[ (\sqrt{2} + 1)^6 = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{2})^{6-k} (1)^k \]

### Step 2: Calculate each term in the expansion
Calculating each term:
- 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 \(k = 5\): \(\binom{6}{5} (\sqrt{2})^1 = 6 \cdot \sqrt{2} = 6\sqrt{2}\)
- For \(k = 6\): \(\binom{6}{6} (1)^6 = 1\)

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

### Step 4: Approximate \(\sqrt{2}\)
Using the approximation \(\sqrt{2} \approx 1.414\):
\[ 70\sqrt{2} \approx 70 \cdot 1.414 \approx 99.98 \]

### Step 5: Calculate the total
Now we calculate:
\[ 99 + 70\sqrt{2} \approx 99 + 99.98 \approx 198.98 \]

### Step 6: Apply the greatest integer function
Now we apply the greatest integer function:
\[ \left\lfloor 198.98 \right\rfloor = 198 \]

### Final Answer
Thus, \(\left\lfloor (\sqrt{2} + 1)^6 \right\rfloor = 198\).