Integral part of `(sqrt2+1)^6` is

Options

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

Answer

To find the integral part of \((\sqrt{2}+1)^6\), we can follow these steps:

### Step 1: Use the Binomial Theorem
According to the Binomial Theorem, we can expand \((a + b)^n\) as follows:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, let \(a = \sqrt{2}\), \(b = 1\), and \(n = 6\):
\[ (\sqrt{2} + 1)^6 = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{2})^{6-k} (1)^k \]

### Step 2: Calculate Each Term
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 \(k = 5\): \(\binom{6}{5} (\sqrt{2})^1 = 6 \cdot \sqrt{2} = 6\sqrt{2}\)
- For \(k = 6\): \(\binom{6}{6} (\sqrt{2})^0 = 1\)

### Step 3: Combine the Terms
Now we can combine the integer and irrational parts:
\[ (\sqrt{2}+1)^6 = 8 + 60 + 30 + 1 + (24\sqrt{2} + 40\sqrt{2} + 6\sqrt{2}) = 99 + 70\sqrt{2} \]

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

### Step 5: Calculate the Total
Now we can add the integer part:
\[ 99 + 99.98 \approx 198.98 \]

### Step 6: Find the Integral Part
The integral part of \(198.98\) is \(198\).

### Conclusion
Thus, the integral part of \((\sqrt{2} + 1)^6\) is:
\[ \boxed{198} \]