If `a_(r)` is the coefficient of `x^(r )` in the expansion of `(1+x+x^(2))^(n)(n in N)`. Then the value of `(a_(1)+4a_(4)+7a_(7)+10a_(10)+……..)` is equal to :

Options

A. a) `3^(n-1)`
B. b) `2^(n)`
C. c) `(1)/(3)*2^(n)`
D. d) `n*3^(n-1)`

Answer

To solve the problem, we need to find the value of the series \( a_1 + 4a_4 + 7a_7 + 10a_{10} + \ldots \) where \( a_r \) is the coefficient of \( x^r \) in the expansion of \( (1 + x + x^2)^n \).

### Step-by-step Solution:

1. **Understand the Expansion**:
The expression \( (1 + x + x^2)^n \) can be expanded using the multinomial theorem. The coefficients \( a_r \) represent the number of ways to select terms from the expansion such that the total exponent of \( x \) equals \( r \).

2. **Identify the Coefficients**:
The coefficients \( a_r \) can be interpreted as the number of non-negative integer solutions to the equation \( x_1 + 2x_2 + 3x_3 = r \), where \( x_1, x_2, x_3 \) are the counts of how many times we take \( 1, x, x^2 \) respectively.

3. **Differentiation Approach**:
To find the sum \( a_1 + 4a_4 + 7a_7 + 10a_{10} + \ldots \), we can differentiate the generating function \( (1 + x + x^2)^n \) with respect to \( x \).

4. **First Derivative**:
Differentiate \( (1 + x + x^2)^n \):
\[ \frac{d}{dx}[(1 + x + x^2)^n] = n(1 + x + x^2)^{n-1}(1 + 2x) \]

5. **Evaluate at Specific Points**:
- Evaluate at \( x = 1 \):
\[ n(1 + 1 + 1)^{n-1}(1 + 2) = n \cdot 3^{n-1} \cdot 3 = 3n \cdot 3^{n-1} \]
This gives us the sum \( a_1 + 2a_2 + 3a_3 + \ldots + 2na_{2n} \).

6. **Using Roots of Unity**:
To isolate the coefficients \( a_1, a_4, a_7, \ldots \), we can use the roots of unity method. Let \( \omega = e^{2\pi i / 3} \), a primitive cube root of unity.
- Evaluate at \( x = \omega \) and \( x = \omega^2 \):
\[ 0 = a_1 + 2a_2\omega + 3a_3\omega^2 + \ldots \]
\[ 0 = a_1 + 2a_2\omega^2 + 3a_3\omega + \ldots \]

7. **Summing the Equations**:
By adding the three equations derived from evaluating at \( x = 1, \omega, \omega^2 \), we can isolate the desired coefficients:
\[ 3n \cdot 3^{n-1} = 3(a_1 + 4a_4 + 7a_7 + \ldots) \]

8. **Final Calculation**:
Dividing both sides by 3 gives:
\[ a_1 + 4a_4 + 7a_7 + \ldots = n \cdot 3^{n-1} \]

### Conclusion:
The value of \( a_1 + 4a_4 + 7a_7 + 10a_{10} + \ldots \) is equal to \( n \cdot 3^{n-1} \).