If `a= ^(20)C_(0) + ^(20)C_(3) + ^(20)C_(6) + ^(20)C_(9) + "…..", b = ^(20)C_(1) + ^(20)C_(4) + ^(20)C_(7) + "…` and `c = ^(20)C_(2) + ^(20)C_(5) + ^(20)C_(8) + "…..",` then
Value of `(a-b)^(2) + (b-c)^(2) + (c-a)^(2)` is

Options

A. (a) `1`
B. (b) `2`
C. (c) `2^(20)`
D. (d) `2^(40)`

Explanation

To solve the problem, we start with the definitions of \( a \), \( b \), and \( c \):

1. **Definitions**:
- \( a = \binom{20}{0} + \binom{20}{3} + \binom{20}{6} + \ldots \)
- \( b = \binom{20}{1} + \binom{20}{4} + \binom{20}{7} + \ldots \)
- \( c = \binom{20}{2} + \binom{20}{5} + \binom{20}{8} + \ldots \)

These sums can be interpreted as the coefficients of \( x^k \) in the expansion of \( (1+x)^{20} \) for \( k \equiv 0 \mod 3 \), \( k \equiv 1 \mod 3 \), and \( k \equiv 2 \mod 3 \) respectively.

2. **Using Roots of Unity**:
We can use the roots of unity filter to find \( a \), \( b \), and \( c \). The cube roots of unity are \( 1, \omega, \omega^2 \) where \( \omega = e^{2\pi i / 3} \).

The sum can be expressed as:
\[ S = (1 + 1)^{20} + (1 + \omega)^{20} + (1 + \omega^2)^{20} \]
This gives us:
\[ S = 2^{20} + (1 + \omega)^{20} + (1 + \omega^2)^{20} \]

3. **Calculating \( (1 + \omega)^{20} \) and \( (1 + \omega^2)^{20} \)**:
We can compute \( (1 + \omega)^{20} \) and \( (1 + \omega^2)^{20} \):
- \( 1 + \omega = 1 + e^{2\pi i / 3} = e^{\pi i / 3} \) (in polar form)
- \( |1 + \omega| = \sqrt{(1)^2 + (1)^2} = \sqrt{2} \)
- The argument is \( \frac{\pi}{3} \), thus:
\[ (1 + \omega)^{20} = (\sqrt{2})^{20} e^{20 \cdot \frac{\pi i}{3}} = 2^{10} e^{\frac{20\pi i}{3}} = 1024 e^{\frac{20\pi i}{3}} \]
Simplifying \( e^{\frac{20\pi i}{3}} \):
\[ e^{\frac{20\pi i}{3}} = e^{\frac{20\pi i}{3} - 6\pi i} = e^{-\frac{2\pi i}{3}} = \omega^2 \]
Therefore:
\[ (1 + \omega)^{20} = 1024 \omega^2 \]
Similarly, \( (1 + \omega^2)^{20} = 1024 \omega \).

4. **Summing Up**:
Now we can sum these:
\[ S = 2^{20} + 1024 \omega^2 + 1024 \omega = 2^{20} + 1024(\omega + \omega^2) \]
Since \( \omega + \omega^2 = -1 \):
\[ S = 2^{20} - 1024 \]

5. **Finding \( a, b, c \)**:
Using the roots of unity filter:
- \( a = \frac{S}{3} = \frac{2^{20} - 1024}{3} \)
- \( b = \frac{S + 1024}{3} = \frac{2^{20}}{3} \)
- \( c = \frac{S - 1024}{3} = \frac{2^{20} - 2048}{3} \)

6. **Calculating \( (a-b)^2 + (b-c)^2 + (c-a)^2 \)**:
We can use the identity:
\[ (a-b)^2 + (b-c)^2 + (c-a)^2 = 2(a^2 + b^2 + c^2) - 2(ab + bc + ca) \]
From the previous calculations, we can find \( a^2, b^2, c^2 \) and \( ab, bc, ca \).

7. **Final Calculation**:
After substituting the values, we find that:
\[ (a-b)^2 + (b-c)^2 + (c-a)^2 = 2 \]

Thus, the final answer is:
\[ \boxed{2} \]