Find $\sum_{k = 0}^{11} \frac{1}{k+1}\binom{11}{k}$

Question

How could I solve this sum using properties from Pascal's Triangle and Pascal's rule? $$ S = \frac{\binom{11}{0}}{1} + \frac{\binom{11}{1}}{2} + \frac{\binom{11}{2}}{3} + \ldots + \frac{\binom{11}{11}}{12}. $$

Answer 1

Hint: Use $$\frac{\binom{n}{k}}{k+1} = \frac{n!}{(k+1)!(n-k)!} = \frac{1}{n+1}\binom{n+1}{k+1}$$ to show $$\sum_{k=0}^{n}\frac{\binom{n}{k}}{k+1} = \frac{2^{n+1}-1}{n+1}.$$

Answer 2

Note that $$ \sum_{k=0}^{11}\frac{1}{k+1}\binom{11}{k} =\sum_{k=0}^{11}\binom{11}{k}\int_0^1x^{k}\, dx= \int_0^1 \sum_{k=0}^{11}\binom{11}{k}x^k\, dx=\int_0^1 (1+x)^{11}\, dx $$ by the binomial theorem. The integral can be easily computed.