If `(1+x)^(n)= sum_(r=0)^(n)C_(r )x^(r ) and sum_(r =0)^(n) (C_(r ))/(r+1)=k` then the value of k is

Options

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

Answer

To solve the problem, we need to find the value of \( k \) given the expression:

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

and

\[ \sum_{r=0}^{n} \frac{C_r}{r+1} = k \]

### Step-by-Step Solution:

1. **Start with the Binomial Expansion**:
The binomial theorem states that:
\[ (1+x)^n = \sum_{r=0}^{n} C_r x^r \]
where \( C_r \) is the binomial coefficient \( \binom{n}{r} \).

2. **Integrate Both Sides**:
We will integrate the binomial expansion from \( 0 \) to \( 1 \):
\[ \int_0^1 (1+x)^n \, dx = \int_0^1 \sum_{r=0}^{n} C_r x^r \, dx \]

3. **Calculate the Left Side**:
The left side can be calculated as:
\[ \int_0^1 (1+x)^n \, dx = \left[ \frac{(1+x)^{n+1}}{n+1} \right]_0^1 = \frac{(1+1)^{n+1}}{n+1} - \frac{(1+0)^{n+1}}{n+1} = \frac{2^{n+1} - 1}{n+1} \]

4. **Calculate the Right Side**:
The right side can be calculated as:
\[ \int_0^1 \sum_{r=0}^{n} C_r x^r \, dx = \sum_{r=0}^{n} C_r \int_0^1 x^r \, dx = \sum_{r=0}^{n} C_r \left[ \frac{x^{r+1}}{r+1} \right]_0^1 = \sum_{r=0}^{n} \frac{C_r}{r+1} \]

5. **Equate Both Sides**:
From the above calculations, we have:
\[ \frac{2^{n+1} - 1}{n+1} = \sum_{r=0}^{n} \frac{C_r}{r+1} \]
Therefore, we can conclude that:
\[ k = \frac{2^{n+1} - 1}{n+1} \]

### Final Result:
Thus, the value of \( k \) is:
\[ \boxed{\frac{2^{n+1} - 1}{n+1}} \]