`sum_(r=1)^(2002+(2k-1))cos((2rpi)/7)+isin((2rpi)/7)=0` then the non negative integtral values of k less than 10 may be

Answer

To solve the problem, we need to evaluate the expression

\[ \sum_{r=1}^{2002 + (2k - 1)} \left( \cos\left(\frac{2r\pi}{7}\right) + i \sin\left(\frac{2r\pi}{7}\right) \right) = 0 \]

This can be rewritten using Euler's formula:

\[ \sum_{r=1}^{2002 + (2k - 1)} e^{i \frac{2r\pi}{7}} = 0 \]

### Step 1: Identify the sum as a geometric series

The expression can be recognized as a geometric series where the first term \( a = e^{i \frac{2\pi}{7}} \) and the common ratio \( r = e^{i \frac{2\pi}{7}} \). The number of terms \( n \) is \( 2002 + (2k - 1) = 2001 + 2k \).

The sum of a geometric series is given by:

\[ S_n = a \frac{1 - r^n}{1 - r} \]

Substituting our values:

\[ S = e^{i \frac{2\pi}{7}} \cdot \frac{1 - \left(e^{i \frac{2\pi}{7}}\right)^{2001 + 2k}}{1 - e^{i \frac{2\pi}{7}}} \]

### Step 2: Set the sum equal to zero

For the sum to equal zero, the numerator must be zero (since the denominator is non-zero):

\[ 1 - \left(e^{i \frac{2\pi}{7}}\right)^{2001 + 2k} = 0 \]

This implies:

\[ \left(e^{i \frac{2\pi}{7}}\right)^{2001 + 2k} = 1 \]

### Step 3: Analyze the condition for equality to 1

The expression \( e^{i \frac{2\pi}{7}} \) raised to any integer multiple of \( 7 \) will equal \( 1 \). Therefore, we need:

\[ \frac{2001 + 2k}{7} = n \quad \text{(where \( n \) is an integer)} \]

This implies that:

\[ 2001 + 2k = 7n \]

### Step 4: Rearranging the equation

Rearranging gives us:

\[ 2k = 7n - 2001 \]

\[ k = \frac{7n - 2001}{2} \]

### Step 5: Finding integer values for \( k \)

For \( k \) to be a non-negative integer, \( 7n - 2001 \) must be non-negative and even. Therefore:

1. \( 7n \geq 2001 \)
2. \( 7n \) must be even.

Since \( 2001 \) is odd, \( 7n \) must also be odd. This means \( n \) must be odd. Let \( n = 2m + 1 \) (where \( m \) is a non-negative integer):

Substituting gives:

\[ 7(2m + 1) - 2001 = 14m + 7 - 2001 = 14m - 1994 \]

Thus:

\[ k = \frac{14m - 1994}{2} = 7m - 997 \]

### Step 6: Finding values of \( k \)

To ensure \( k \) is non-negative:

\[ 7m - 997 \geq 0 \implies 7m \geq 997 \implies m \geq \frac{997}{7} \approx 142.43 \]

Thus, \( m \) must be at least \( 143 \).

### Step 7: Finding the upper limit for \( k \)

Since \( k < 10 \):

\[ 7m - 997 < 10 \implies 7m < 1007 \implies m < \frac{1007}{7} \approx 143.86 \]

Thus, \( m \) can only take the value \( 143 \).

### Step 8: Calculate \( k \)

Substituting \( m = 143 \):

\[ k = 7(143) - 997 = 1001 - 997 = 4 \]

### Conclusion

The only non-negative integral value of \( k \) less than \( 10 \) is:

\[ \boxed{4} \]