The sum of series `3*""^(2007)C_(0)-8*""^(2007)C_(1)+13*""^(2007)C_(2)-18*""^(2007)C_(3)+……` upto 2008 terms is K, then K is :

Answer

To find the sum of the series \( S = 3 \cdot \binom{2007}{0} - 8 \cdot \binom{2007}{1} + 13 \cdot \binom{2007}{2} - 18 \cdot \binom{2007}{3} + \ldots \) up to 2008 terms, we can express the series in a more manageable form.

### Step 1: Identify the general term
The coefficients \( 3, -8, 13, -18, \ldots \) can be expressed as a linear function of \( n \). We observe that:
- The first term is \( 3 \) (when \( n = 0 \)),
- The second term is \( -8 \) (when \( n = 1 \)),
- The third term is \( 13 \) (when \( n = 2 \)),
- The fourth term is \( -18 \) (when \( n = 3 \)).

We can see that the coefficients follow an arithmetic progression (AP) with a common difference of \( 5 \). The general term can be expressed as:
\[ a_n = 5n + 3 \]
However, since the signs alternate, we will include \( (-1)^n \) in our expression.

### Step 2: Write the series in summation notation
The series can be rewritten using summation notation:
\[ S = \sum_{n=0}^{2007} (-1)^n (5n + 3) \cdot \binom{2007}{n} \]

### Step 3: Split the summation
We can split the summation into two parts:
\[ S = \sum_{n=0}^{2007} (-1)^n (5n) \cdot \binom{2007}{n} + \sum_{n=0}^{2007} (-1)^n 3 \cdot \binom{2007}{n} \]
This simplifies to:
\[ S = 5 \sum_{n=0}^{2007} (-1)^n n \cdot \binom{2007}{n} + 3 \sum_{n=0}^{2007} (-1)^n \binom{2007}{n} \]

### Step 4: Evaluate the second summation
The second summation can be evaluated using the binomial theorem:
\[ \sum_{n=0}^{2007} (-1)^n \binom{2007}{n} = (1 - 1)^{2007} = 0 \]

### Step 5: Evaluate the first summation
For the first summation, we use the identity:
\[ \sum_{n=0}^{k} n \cdot \binom{k}{n} = k \cdot 2^{k-1} \]
Thus,
\[ \sum_{n=0}^{2007} n \cdot \binom{2007}{n} = 2007 \cdot 2^{2006} \]
Now, incorporating the alternating sign:
\[ \sum_{n=0}^{2007} (-1)^n n \cdot \binom{2007}{n} = -2007 \cdot 2^{2006} \]

### Step 6: Combine the results
Putting it all together:
\[ S = 5(-2007 \cdot 2^{2006}) + 3(0) = -10035 \cdot 2^{2006} \]

### Final Result
Thus, the sum of the series \( K \) is:
\[ K = -10035 \cdot 2^{2006} \]