If `(x+1/x+1)^(6)=a_(0)+(a_(1)x+(b_(1))/(x))+(a_(2)x^(2)+(b_(2))/(x^(2)))+"...."+(a_(6)x^(6)+(b_(6))/(x^(6)))`, then

Options

A. `a_() = 141`
B. `a_(5) = 6`
C. `underset(i=1)overset(6)sum a_(i) + b_(i) = 588`
D. `underset(i=1)overset(6)suma_(i) + b_(i) = 3^(6)`

Explanation

To solve the problem, we need to expand the expression \((x + \frac{1}{x} + 1)^6\) using the Binomial Theorem and identify the coefficients \(a_i\) and \(b_i\) for the given polynomial form.

### Step-by-Step Solution:

1. **Rewrite the Expression**:
\[ (x + \frac{1}{x} + 1)^6 \]

2. **Identify the Terms**:
We can treat \(x + \frac{1}{x} + 1\) as a single term. To apply the Binomial Theorem, we can consider it as:
\[ (a + b + c)^n \text{ where } a = x, b = \frac{1}{x}, c = 1, n = 6 \]

3. **Apply the Multinomial Expansion**:
The multinomial expansion gives us:
\[ \sum_{i+j+k=6} \frac{6!}{i!j!k!} x^i \left(\frac{1}{x}\right)^j 1^k \]
where \(i\), \(j\), and \(k\) are non-negative integers.

4. **Simplify the Terms**:
Each term in the expansion can be simplified to:
\[ \frac{6!}{i!j!k!} x^{i-j} \]
The term \(x^{i-j}\) indicates that the power of \(x\) will depend on the values of \(i\) and \(j\).

5. **Find Coefficients**:
We need to find the coefficients \(a_i\) and \(b_i\) for \(x^i\) and \(\frac{1}{x^j}\). The coefficients can be derived from the multinomial expansion.

6. **Calculate \(a_0\)**:
To find \(a_0\), we need the constant term when \(x = 1\):
\[ (1 + 1 + 1)^6 = 3^6 = 729 \]
The constant term \(a_0\) can be calculated by considering the contributions from all terms where \(i = j = 0\).

7. **Calculate \(a_5\)**:
For \(a_5\), we need the coefficient of \(x^5\):
\[ \text{Coefficient of } x^5 = \text{terms where } i = 5, j = 0, k = 1 \]
This gives us:
\[ \frac{6!}{5!0!1!} = 6 \]

8. **Calculate the Summation**:
To find the summation \(\sum_{i=1}^{6} (a_i + b_i)\), we can substitute \(x = 1\) into the expanded polynomial:
\[ 3^6 - a_0 = 729 - 141 = 588 \]

9. **Final Result**:
Thus, the final answer for the summation of \(a_i + b_i\) from \(i = 1\) to \(6\) is:
\[ \sum_{i=1}^{6} (a_i + b_i) = 588 \]