If `(1+x)^(2010)=C_(0)+C_(1)x+C_(2)x^(2)+ …….+C_(2010) x^(2010)` then the sum of series `C_(2)+C_(5)+C_(8)+ ……… +C_(2009)` equals to :

Options

A. `(1)/(2)(2^(2010)-1)`
B. `(1)/(3) (2^(2010)-1)`
C. `(1)/(2) (2^(2009)-1)`
D. `(1)/(3)(2^(2009)-1)`

Answer

To solve the problem of finding the sum of the series \( C_2 + C_5 + C_8 + \ldots + C_{2009} \) from the expansion of \( (1+x)^{2010} = C_0 + C_1 x + C_2 x^2 + \ldots + C_{2010} x^{2010} \), we can use the properties of roots of unity.

### Step-by-Step Solution:

1. **Understanding the Coefficients**:
The coefficients \( C_k \) in the expansion of \( (1+x)^{2010} \) are given by \( C_k = \binom{2010}{k} \).

2. **Identifying the Series**:
We need to find the sum \( C_2 + C_5 + C_8 + \ldots + C_{2009} \). This series consists of coefficients where the indices are of the form \( 2 + 3k \) for \( k = 0, 1, 2, \ldots \).

3. **Using Roots of Unity**:
To extract the coefficients where the index is of the form \( 2 + 3k \), we can use the roots of unity filter. Let \( \omega = e^{2\pi i / 3} \) be a primitive cube root of unity. The sum can be calculated using:
\[ S = \frac{1}{3} \left( (1+1)^{2010} + (1+\omega)^{2010} + (1+\omega^2)^{2010} \right) \]

4. **Calculating Each Term**:
- For the first term:
\[ (1+1)^{2010} = 2^{2010} \]
- For the second term:
\[ 1 + \omega = 2 \cos\left(\frac{\pi}{3}\right) = 1 \quad \Rightarrow \quad (1+\omega)^{2010} = 1^{2010} = 1 \]
- For the third term:
\[ 1 + \omega^2 = 2 \cos\left(-\frac{\pi}{3}\right) = 1 \quad \Rightarrow \quad (1+\omega^2)^{2010} = 1^{2010} = 1 \]

5. **Summing the Results**:
Now substituting back into the equation for \( S \):
\[ S = \frac{1}{3} \left( 2^{2010} + 1 + 1 \right) = \frac{1}{3} \left( 2^{2010} + 2 \right) = \frac{2^{2010} + 2}{3} \]

6. **Final Result**:
Thus, the sum \( C_2 + C_5 + C_8 + \ldots + C_{2009} \) is:
\[ \frac{2^{2010} + 2}{3} \]