The sum of the series `1+(1)/(2) ""^(n) C_1 + (1)/(3) ""^(n) C_(2) + ….+ (1)/(n+1) ""^(n) C_(n)` is equal to

Options

A. `(2^(n+1) -1)/(n+1)`
B. `(3(2^(n)-1))/(2n)`
C. `^(2^(n)+1)/(n+1)`
D. `(2^(n) +1)/(2n)`

Answer

To solve the series

\[ S = 1 + \frac{1}{2} \binom{n}{1} + \frac{1}{3} \binom{n}{2} + \ldots + \frac{1}{n+1} \binom{n}{n}, \]

we can use the concept of integration in relation to binomial coefficients.

### Step 1: Recognize the Binomial Expansion

The binomial expansion of \((1 + x)^n\) is given by:

\[ (1 + x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r. \]

### Step 2: Integrate the Binomial Expansion

To incorporate the denominators in our series, we can integrate both sides of the binomial expansion from \(0\) to \(1\):

\[ \int_0^1 (1 + x)^n \, dx = \int_0^1 \sum_{r=0}^{n} \binom{n}{r} x^r \, dx. \]

### Step 3: Calculate the Left-Hand Side

The left-hand side can be computed as follows:

\[ \int_0^1 (1 + x)^n \, dx = \left[ \frac{(1 + x)^{n+1}}{n + 1} \right]_0^1 = \frac{(1 + 1)^{n+1}}{n + 1} - \frac{(1 + 0)^{n+1}}{n + 1} = \frac{2^{n+1}}{n + 1} - \frac{1}{n + 1}. \]

This simplifies to:

\[ \frac{2^{n+1} - 1}{n + 1}. \]

### Step 4: Calculate the Right-Hand Side

Now, we calculate the right-hand side:

\[ \int_0^1 \sum_{r=0}^{n} \binom{n}{r} x^r \, dx = \sum_{r=0}^{n} \binom{n}{r} \int_0^1 x^r \, dx = \sum_{r=0}^{n} \binom{n}{r} \cdot \frac{1}{r + 1}. \]

This means:

\[ \sum_{r=0}^{n} \frac{\binom{n}{r}}{r + 1} = S. \]

### Step 5: Equate Both Sides

We can now equate the two sides:

\[ S = \frac{2^{n+1} - 1}{n + 1}. \]

### Final Result

Thus, the sum of the series is:

\[ \boxed{\frac{2^{n+1} - 1}{n + 1}}. \]