Find the number of solution of the equation `underset(l=1)overset(501)Sigma x_(l)=1501`, where `x_(l)'s` are odd natural numbers:

Answer

To find the number of solutions to the equation

\[ \sum_{l=1}^{501} x_l = 1501 \]

where \( x_l \) are odd natural numbers, we can follow these steps:

### Step 1: Express odd natural numbers
Odd natural numbers can be expressed in the form \( x_l = 2n_l + 1 \), where \( n_l \) is a non-negative integer. This means that each \( x_l \) can be rewritten as:

\[ x_1 = 2n_1 + 1, \quad x_2 = 2n_2 + 1, \quad \ldots, \quad x_{501} = 2n_{501} + 1 \]

### Step 2: Substitute into the equation
Substituting these expressions into the original equation gives:

\[ (2n_1 + 1) + (2n_2 + 1) + \ldots + (2n_{501} + 1) = 1501 \]

### Step 3: Simplify the equation
This can be simplified as follows:

\[ 2(n_1 + n_2 + \ldots + n_{501}) + 501 = 1501 \]

### Step 4: Isolate the sum of \( n_l \)
Now, isolate the sum of \( n_l \):

\[ 2(n_1 + n_2 + \ldots + n_{501}) = 1501 - 501 \]

This simplifies to:

\[ 2(n_1 + n_2 + \ldots + n_{501}) = 1000 \]

### Step 5: Divide by 2
Dividing both sides by 2 gives:

\[ n_1 + n_2 + \ldots + n_{501} = 500 \]

### Step 6: Find the number of non-negative integer solutions
Now, we need to find the number of non-negative integer solutions to the equation:

\[ n_1 + n_2 + \ldots + n_{501} = 500 \]

This is a classic problem in combinatorics, and the number of solutions can be found using the "stars and bars" theorem. The formula for the number of solutions in non-negative integers to the equation

\[ x_1 + x_2 + \ldots + x_r = n \]

is given by:

\[ \binom{n + r - 1}{r - 1} \]

### Step 7: Apply the formula
In our case, \( n = 500 \) and \( r = 501 \):

\[ \text{Number of solutions} = \binom{500 + 501 - 1}{501 - 1} = \binom{1000}{500} \]

### Final Answer
Thus, the number of solutions to the equation is:

\[ \binom{1000}{500} \]

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