Explanation:

To find the sum of coefficients in the expansion of (x + 2y + z)^n, we can use the Binomial Theorem.

The Binomial Theorem states that for any positive integer n, the expansion of (x + y)^n can be written as:

(x + y)^n = C(n, 0)x^n y^0 + C(n, 1)x^(n-1) y^1 + C(n, 2)x^(n-2) y^2 + ... + C(n, n-1)x^1 y^(n-1) + C(n, n)x^0 y^n

Where C(n, k) is the binomial coefficient, given by:

C(n, k) = n! / (k!(n-k)!)

In the case of (x + 2y + z)^n, the coefficients will be found by replacing x with 1, y with 2, and z with 1. Therefore, the expansion becomes:

(1 + 2(2y) + 1)^n = (1 + 4y + 1)^n

Now we can apply the Binomial Theorem to find the coefficients. The sum of the coefficients will be the sum of the terms in the expansion.

Applying the Binomial Theorem:

Using the Binomial Theorem, the expansion of (1 + 4y + 1)^n can be written as:

(1 + 4y + 1)^n = C(n, 0)(1)^n (4y)^0 + C(n, 1)(1)^(n-1) (4y)^1 + C(n, 2)(1)^(n-2) (4y)^2 + ... + C(n, n-1)(1)^1 (4y)^(n-1) + C(n, n)(1)^0 (4y)^n

Simplifying this expression, we get:

(1 + 4y + 1)^n = C(n, 0) + 4C(n, 1)y + 6C(n, 2)(4y)^2 + ... + nC(n, n-1)(4y)^(n-1) + C(n, n)(4y)^n

The sum of the coefficients will be the sum of the terms in this expansion.

Calculating the Sum of the Coefficients:

To calculate the sum of the coefficients, we need to sum up all the terms in the expansion.

The sum of the coefficients can be written as:

C(n, 0) + 4C(n, 1) + 6C(n, 2) + ... + nC(n, n-1) + C(n, n)

Each term in this sum is a binomial coefficient multiplied by a constant factor.

Conclusion:

In general, the sum of the coefficients in the expansion of (x + 2y + z)^n is not equal to any of the given options (a, b, c). Therefore, the correct answer is option D, "none of these".