If `(1+x)^(n)=C_(0)+C_(1)x+…..+C_(n)x^(n)`, then `(C_(1))/(C_(0))+(2C_(2))/(C_(1))+(3C_(3))/(C_(2))+....+(nC_(n))/(C_(n-1))` is :

Options

A. `(2^(n))/(n!)`
B. `(n+1)^(n)/(n!)`
C. `(n(n-1))/(2)`
D. `(n(n+1))/(2)`

Answer

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

\[ \frac{C_1}{C_0} + \frac{2C_2}{C_1} + \frac{3C_3}{C_2} + \ldots + \frac{nC_n}{C_{n-1}} \]

where \( C_k = \binom{n}{k} \) are the binomial coefficients from the expansion of \( (1+x)^n \).

### Step 1: Rewrite the Expression

We can rewrite the expression in summation notation:

\[ S = \sum_{r=1}^{n} \frac{r C_r}{C_{r-1}} \]

### Step 2: Substitute Binomial Coefficients

Using the property of binomial coefficients, we know:

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

Thus, we can express \( \frac{C_r}{C_{r-1}} \):

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

### Step 3: Substitute Back into the Summation

Now substituting this back into our summation:

\[ S = \sum_{r=1}^{n} r \cdot \frac{n-r+1}{r} = \sum_{r=1}^{n} (n - r + 1) \]

### Step 4: Simplify the Summation

Now we can simplify the summation:

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

This can be rewritten as:

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

The first summation \( \sum_{r=1}^{n} 1 = n \) and the second summation \( \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \).

### Step 5: Combine the Results

Putting it all together:

\[ S = (n + 1)n - \frac{n(n + 1)}{2} \]

This simplifies to:

\[ S = n(n + 1) - \frac{n(n + 1)}{2} = \frac{2n(n + 1) - n(n + 1)}{2} = \frac{n(n + 1)}{2} \]

### Final Result

Thus, the final result is:

\[ S = \frac{n(n + 1)}{2} \]

### Conclusion

The expression evaluates to:

\[ \frac{n(n + 1)}{2} \]