Question:

<p><strong>Step 1:</strong> Start with the binomial expansion: $(1+x)^n = \sum_{r=0}^{n} {}^nC_r x^r$</p><p><strong>Step 2:</strong> Integrate both sides from 0 to 1:</p><p>$\int_0^1 (1+x)^n dx = \int_0^1 \sum_{r=0}^{n} {}^nC_r x^r dx = \sum_{r=0}^{n} {}^nC_r \int_0^1 x^r dx$</p><p><strong>Step 3:</strong> Evaluate the left side: $\int_0^1 (1+x)^n dx = \left[\frac{(1+x)^{n+1}}{n+1}\right]_0^1 = \frac{2^{n+1}-1}{n+1}$</p><p><strong>Step 4:</strong> Evaluate the right side: $\sum_{r=0}^{n} {}^nC_r \cdot \frac{1}{r+1} = \sum_{r=0}^{n} \frac{{}^nC_r}{r+1}$</p><p><strong>Step 5:</strong> Now use $(1-x)^n$ for the alternating sum. Integrate $(1-x)^n$ from 0 to 1:</p><p>$\int_0^1 (1-x)^n dx = \left[-\frac{(1-x)^{n+1}}{n+1}\right]_0^1 = \frac{1}{n+1}$</p><p><strong>Step 6:</strong> This gives: $\sum_{r=0}^{n} {}^nC_r(-1)^r \cdot \frac{1}{r+1} = \frac{1}{n+1}$</p><p><strong>Step 7:</strong> Separating r=0 term: ${}^nC_0 \cdot 1 + \sum_{r=1}^{n}(-1)^r\frac{{}^nC_r}{r+1} = \frac{1}{n+1}$</p><p><strong>Step 8:</strong> Therefore: $\sum_{r=1}^{n}(-1)^{r+1}\frac{{}^nC_r}{r+1} = -\left(\frac{1}{n+1}-1\right) = \frac{n}{n+1}$</p><p>∴ Answer: C</p>