If `C_(0), C_(1), C_(2),……, C_(20)` are the binomial coefficients in the expansion of `(1+x)^(20)`, then the value of `(C_(1))/(C_(0))+2(C_(2))/(C_(1))+3(C_(3))/(C_(2))+……+19(C_(19))/(C_(18))+20(C_(20))/(C_(19))` is equal to (where `C_(r)` represetns `.^(n)C_(r)`)

Options

A. 120
B. 210
C. 180
D. 240

Answer

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

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

where \( C_r = \binom{20}{r} \).

### Step 1: Express each term in the sum

We know that the binomial coefficients can be expressed as:

\[ C_r = \binom{20}{r} \quad \text{and} \quad C_{r-1} = \binom{20}{r-1} \]

Thus, we can rewrite each term in the sum:

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

### Step 2: Substitute back into the sum

Now substituting this back into the sum \( S \):

\[ S = \sum_{r=1}^{20} r \cdot \frac{C_r}{C_{r-1}} = \sum_{r=1}^{20} r \cdot \frac{20-r+1}{r} = \sum_{r=1}^{20} (20 - r + 1) = \sum_{r=1}^{20} (21 - r) \]

### Step 3: Evaluate the sum

Now we can evaluate the sum:

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

The first part is:

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

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

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

### Step 4: Combine results

Now substituting back into the expression for \( S \):

\[ S = 420 - 210 = 210 \]

Thus, the value of the expression is:

\[ \boxed{210} \]