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. `3^(n-1)`
B. `2^(n)`
C. `(1)/(3)*2^(n)`
D. `n*3^(n-1)`

Answer

To find the value of \( a_1 + 4a_4 + 7a_7 + 10a_{10} + \ldots \) where \( a_r \) is the coefficient of \( x^r \) in the expansion of \( (1 + x + x^2)^n \), we can follow these steps:

### Step 1: Understanding the Coefficients
The expression \( (1 + x + x^2)^n \) can be expanded using the multinomial theorem. The coefficient \( a_r \) represents the number of ways to choose terms from \( (1 + x + x^2) \) such that the total degree of \( x \) is \( r \).

### Step 2: Generating Function
We can express \( (1 + x + x^2)^n \) as a generating function. 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 \( x_1 + x_2 + x_3 = n \).

### Step 3: Finding the Required Sum
We need to find \( a_1 + 4a_4 + 7a_7 + 10a_{10} + \ldots \). This can be interpreted as:
\[ \sum_{k=0}^{\infty} (3k + 1) a_{3k + 1} \]
This sum can be computed using the generating function approach.

### Step 4: Differentiating the Generating Function
To find \( a_1 + 4a_4 + 7a_7 + \ldots \), we can differentiate the generating function:
\[ f(x) = (1 + x + x^2)^n \]
Differentiating gives:
\[ f'(x) = n(1 + x + x^2)^{n-1}(1 + 2x) \]
We can evaluate this at \( x = 1 \) to find the sum of coefficients.

### Step 5: Evaluating at \( x = 1 \)
Substituting \( x = 1 \):
\[ f(1) = (1 + 1 + 1)^n = 3^n \]
And differentiating:
\[ f'(1) = n(3^{n-1})(1 + 2) = 3n \cdot 3^{n-1} \]

### Step 6: Using Roots of Unity
To isolate the coefficients \( a_r \), we can use the roots of unity. Let \( \omega = e^{2\pi i / 3} \) be a primitive cube root of unity. We can find:
\[ S = f(1) + f(\omega) + f(\omega^2) \]
This helps in isolating the coefficients.

### Step 7: Final Calculation
Using the results from the previous steps, we find:
\[ S = 3^n + 0 + 0 = 3^n \]
Thus, the sum \( a_1 + 4a_4 + 7a_7 + \ldots \) can be expressed as:
\[ \frac{n \cdot 3^n}{3} = n \cdot 3^{n-1} \]

### Conclusion
The final result is:
\[ \boxed{n \cdot 3^{n-1}} \]