Let `((n),(k))` represents the combination of `'n'` things taken `'k'` at a time, then the value of the sum `((99),(97))+((98),(96))+((97),(95))+.........+((3),(1))+((2),(0))` equals-

Options

A. `((99),(97))`
B. `((100),(98))`
C. `((99),(98))`
D. `((100),(97))`

Answer

To solve the problem, we need to evaluate the sum:

\[ \sum_{k=0}^{2} \binom{99-k}{97-k} = \binom{99}{97} + \binom{98}{96} + \binom{97}{95} + \ldots + \binom{3}{1} + \binom{2}{0} \]

### Step 1: Rewrite the combinations
We can express the combinations in a more manageable form. The combination \(\binom{n}{r}\) can be rewritten as:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

Thus, we can express each term in the sum:

\[ \binom{99}{97} = \frac{99!}{97! \cdot 2!}, \quad \binom{98}{96} = \frac{98!}{96! \cdot 2!}, \quad \ldots, \quad \binom{2}{0} = \frac{2!}{0! \cdot 2!} \]

### Step 2: Factor out the common denominator
Notice that each term has a common denominator of \(2!\). We can factor this out:

\[ \sum_{k=0}^{2} \binom{99-k}{97-k} = \frac{1}{2!} \left( \frac{99!}{97!} + \frac{98!}{96!} + \ldots + \frac{2!}{0!} \right) \]

### Step 3: Simplify the factorial expressions
Now, we can simplify each term:

\[ \frac{99!}{97!} = 99 \cdot 98, \quad \frac{98!}{96!} = 98, \quad \frac{97!}{95!} = 97 \cdot 96, \ldots, \frac{3!}{1!} = 3 \cdot 2 \]

### Step 4: Recognize the pattern
The sum can be recognized as a telescoping series. Each term can be expressed as:

\[ \sum_{k=0}^{97} (99-k)(98-k) \]

This can be simplified using the hockey-stick identity in combinatorics.

### Step 5: Apply the hockey-stick identity
Using the hockey-stick identity:

\[ \sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1} \]

We can evaluate the sum:

\[ \sum_{k=0}^{2} \binom{99-k}{2} = \binom{100}{3} \]

### Step 6: Calculate the final result
Now we can compute \(\binom{100}{3}\):

\[ \binom{100}{3} = \frac{100 \cdot 99 \cdot 98}{3 \cdot 2 \cdot 1} = \frac{970200}{6} = 161700 \]

### Final Answer
Thus, the value of the sum is:

\[ \boxed{161700} \]