`(C_(1))/(C_(0))+(2.C_(2))/(C_(1))+(3.C_(3))/(C_(2))+ . . .+(20.C_(20))/(C_(19))=`

Options

A. 180
B. 210
C. 240
D. 280

Answer

To solve the given expression

\[ S = \frac{C_1}{C_0} + \frac{2 \cdot C_2}{C_1} + \frac{3 \cdot C_3}{C_2} + \ldots + \frac{20 \cdot C_{20}}{C_{19}}, \]

we will break it down step by step.

### Step 1: Understanding the terms
The general term in the series can be expressed as:

\[ T_r = r \cdot \frac{C_r}{C_{r-1}}. \]

Here, \(C_r\) represents the binomial coefficient \(\binom{20}{r}\).

### Step 2: Expressing the binomial coefficients
Using the definition of binomial coefficients, we have:

\[ C_r = \frac{20!}{r!(20-r)!} \quad \text{and} \quad C_{r-1} = \frac{20!}{(r-1)!(20-r+1)!}. \]

### Step 3: Simplifying the ratio
Now, we can simplify the term \(T_r\):

\[ \frac{C_r}{C_{r-1}} = \frac{\frac{20!}{r!(20-r)!}}{\frac{20!}{(r-1)!(20-r+1)!}} = \frac{(20-r+1)}{r}. \]

Thus, we can rewrite \(T_r\):

\[ T_r = r \cdot \frac{(20-r+1)}{r} = 20 - r + 1 = 21 - r. \]

### Step 4: Summing the series
Now, we need to sum \(T_r\) from \(r = 1\) to \(r = 20\):

\[ S = \sum_{r=1}^{20} (21 - r). \]

This can be rewritten as:

\[ S = \sum_{r=1}^{20} 21 - \sum_{r=1}^{20} r. \]

### Step 5: Calculating the sums
The first sum is straightforward:

\[ \sum_{r=1}^{20} 21 = 21 \cdot 20 = 420. \]

The second sum is the sum of the first 20 natural numbers:

\[ \sum_{r=1}^{20} r = \frac{20 \cdot (20 + 1)}{2} = \frac{20 \cdot 21}{2} = 210. \]

### Step 6: Final calculation
Now substituting back, we have:

\[ S = 420 - 210 = 210. \]

### Final Answer
Thus, the final answer is:

\[ \boxed{210}. \]