Step 1: Understanding the Concept:
This problem involves sums of binomial coefficients. The division by integers \( (r+1) \) suggests integration of the binomial expansion. : Key Formula or Approach:
Identity: \( \frac{{}^nC_r}{r+1} = \frac{{}^{n+1}C_{r+1}}{n+1} \). Step 2: Detailed Explanation:
The given series is:
\[ S = \sum_{r=1}^{n} \frac{{}^nC_r}{r+1} \]
Apply the identity \( \frac{{}^nC_r}{r+1} = \frac{{}^{n+1}C_{r+1}}{n+1} \):
\[ S = \sum_{r=1}^{n} \frac{{}^{n+1}C_{r+1}}{n+1} \]
Factor out \( \frac{1}{n+1} \):
\[ S = \frac{1}{n+1} \left[ {}^{n+1}C_2 + {}^{n+1}C_3 + \dots + {}^{n+1}C_{n+1} \right] \]
We know that the total sum of coefficients for power \( (n+1) \) is:
\[ {}^{n+1}C_0 + {}^{n+1}C_1 + {}^{n+1}C_2 + \dots + {}^{n+1}C_{n+1} = 2^{n+1} \]
So, the required bracket sum is:
\[ \text{Sum} = 2^{n+1} - ({}^{n+1}C_0 + {}^{n+1}C_1) \]
\[ \text{Sum} = 2^{n+1} - (1 + n+1) = 2^{n+1} - n - 2 \]
Substitute back:
\[ S = \frac{2^{n+1} - n - 2}{n+1} \]. Step 3: Final Answer:
The sum is \( \frac{2^{n+1}-n-2}{n+1} \).