1.∑r=011(11r)r+1\displaystyle \sum_{r=0}^{11} \frac{\binom{11}{r}}{r+1}r=0∑11r+1(r11) equals:a.21112\frac{2^{11}}{12}12211b.21212\frac{2^{12}}{12}12212c.211−112\frac{2^{11} - 1}{12}12211−1d.212−112\frac{2^{12} - 1}{12}12212−1Login to continueOnly logged in users canattempt or see the solution.
