If (1 + x + x2)n = a0 + a1x + a2x2 + ..... + a2nx2n, then a0 + a3 + a6...
The given expression can be rewritten as:
(1 + x + x^2)^n = a0 + a1x + a2x^2 + ... + a2nx^(2n)
To find the terms a0, a3, a6, ..., we need to find the terms with powers of x that are multiples of 3.
When expanding (1 + x + x^2)^n, the powers of x will be of the form (x^k)(x^l)(x^m), where k + l + m = 2n.
If we want the power of x to be a multiple of 3, we need k, l, and m to all be multiples of 3.
Since k + l + m = 2n, and we want k, l, and m to be multiples of 3, we can let k = 3a, l = 3b, and m = 3c, where a, b, and c are integers.
Substituting these values into k + l + m = 2n, we get 3a + 3b + 3c = 2n.
Simplifying, we have a + b + c = (2n)/3.
Since a, b, and c are integers, (2n)/3 must be an integer.
Therefore, for the powers of x to be multiples of 3, n must be a multiple of 3.
So, the terms a0, a3, a6, ... are the coefficients of x^0, x^3, x^6, ..., respectively.
Since n must be a multiple of 3, we can write n as 3k, where k is an integer.
Then, the terms a0, a3, a6, ... are the coefficients of x^0, x^3, x^6, ..., respectively, when expanding (1 + x + x^2)^(3k).
Therefore, a0, a3, a6, ... = 3k.