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 binomial expansion of \( (1+x)^{2010} \), we can follow these steps:

### Step 1: Understand the Binomial Expansion
The binomial expansion of \( (1+x)^{2010} \) is given by:
\[ (1+x)^{2010} = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \ldots + C_{2010} x^{2010} \]
where \( C_k = \binom{2010}{k} \).

### Step 2: Multiply the Expansion by \( x \)
To isolate the coefficients \( C_2, C_5, C_8, \ldots \), we can multiply the entire expansion by \( x \):
\[ x(1+x)^{2010} = C_0 x + C_1 x^2 + C_2 x^3 + C_3 x^4 + \ldots + C_{2010} x^{2011} \]

### Step 3: Substitute Roots of Unity
To extract the coefficients \( C_2, C_5, C_8, \ldots \) systematically, we can use the roots of unity. Let \( \omega = e^{2\pi i / 3} \) be a primitive cube root of unity. We will evaluate the expression at \( x = 1 \), \( x = \omega \), and \( x = \omega^2 \).

### Step 4: Evaluate at \( x = 1 \)
Substituting \( x = 1 \):
\[ 1(1+1)^{2010} = 2^{2010} = C_0 + C_1 + C_2 + C_3 + \ldots + C_{2010} \]

### Step 5: Evaluate at \( x = \omega \)
Substituting \( x = \omega \):
\[ \omega(1+\omega)^{2010} = \omega \cdot (1 + e^{2\pi i / 3})^{2010} = \omega \cdot (0)^{2010} = 0 \]
This gives:
\[ C_0 \omega + C_1 \omega^2 + C_2 \omega^3 + C_3 \omega^4 + \ldots = 0 \]

### Step 6: Evaluate at \( x = \omega^2 \)
Substituting \( x = \omega^2 \):
\[ \omega^2(1+\omega^2)^{2010} = \omega^2 \cdot (1 + e^{-2\pi i / 3})^{2010} = \omega^2 \cdot (0)^{2010} = 0 \]
This gives:
\[ C_0 \omega^2 + C_1 \omega + C_2 \omega^2 + C_3 \omega^3 + \ldots = 0 \]

### Step 7: Combine the Results
Now we have three equations:
1. \( 2^{2010} = C_0 + C_1 + C_2 + C_3 + \ldots \)
2. \( 0 = C_0 \omega + C_1 \omega^2 + C_2 \cdot 1 + C_3 \omega + \ldots \)
3. \( 0 = C_0 \omega^2 + C_1 \cdot 1 + C_2 \omega + C_3 \omega^2 + \ldots \)

### Step 8: Solve for the Desired Sum
Adding these equations, we can isolate the terms corresponding to \( C_2, C_5, C_8, \ldots \):
\[ C_2 + C_5 + C_8 + \ldots = \frac{1}{3} (2^{2010} - 1) \]

### Final Answer
Thus, the sum \( C_2 + C_5 + C_8 + \ldots + C_{2009} \) equals:
\[ \frac{1}{3} (2^{2010} - 1) \]