The value of `sum_(r=1)^(10) r. (""^(n)C_(r))/(""^(n)C_(r-1)` is equal to

Options

A. ` 5 (2n -9)`
B. `10`n
C. `9 (n-4)`
D. none of these

Explanation

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

\[ S_n = \sum_{r=1}^{10} r \cdot \frac{nC_r}{nC_{r-1}} \]

### Step 1: Simplify the Binomial Coefficient Ratio
We know that:

\[ \frac{nC_r}{nC_{r-1}} = \frac{n!}{r!(n-r)!} \cdot \frac{(r-1)!(n-(r-1))!}{n!} = \frac{n - r + 1}{r} \]

This means we can rewrite our sum as:

\[ S_n = \sum_{r=1}^{10} r \cdot \frac{n - r + 1}{r} \]

### Step 2: Cancel Out \( r \)
The \( r \) in the numerator and denominator cancels out:

\[ S_n = \sum_{r=1}^{10} (n - r + 1) \]

### Step 3: Rewrite the Summation
Now we can rewrite the summation:

\[ S_n = \sum_{r=1}^{10} (n + 1 - r) \]

### Step 4: Split the Summation
We can split the summation into two parts:

\[ S_n = \sum_{r=1}^{10} (n + 1) - \sum_{r=1}^{10} r \]

### Step 5: Calculate Each Part
1. The first part is simply:

\[ \sum_{r=1}^{10} (n + 1) = 10(n + 1) \]

2. The second part is the sum of the first 10 natural numbers:

\[ \sum_{r=1}^{10} r = \frac{10 \cdot (10 + 1)}{2} = 55 \]

### Step 6: Combine the Results
Now we can combine the results:

\[ S_n = 10(n + 1) - 55 \]

### Step 7: Simplify the Expression
Simplifying gives us:

\[ S_n = 10n + 10 - 55 = 10n - 45 \]

### Step 8: Factor the Expression
We can factor this expression:

\[ S_n = 5(2n - 9) \]

Thus, the final answer is:

\[ S_n = 5(2n - 9) \]

### Final Answer
The value of \( S_n \) is \( 5(2n - 9) \).

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