Question

If n is an odd natural number , then $\sum _{r=0}^n\frac{(-1{)}^r}{{}^r{C}_r}$ equals :

• A. 1/n
• B. n/2ⁿ
• C. 0
• D. None

Answer 1

The correct option is C

0

Let n = 2k+1 then $\sum _{r=0}^n\frac{(-1{)}^r}{{}^n{C}_r}=\sum _{r=0}^{2k+1}\frac{(-1{)}^{2k+1}}{{}^{2k+1}{C}_r}$

= $\sum _{r=0}^n\frac{(-1{)}^r}{{}^n{C}_r}=\sum _{r=0}^{2k+1}\frac{(-1{)}^{2k+1}}{{}^{2k+1}{C}_r}=\frac{1}{{}^{2k+1}{C}_r}-\frac{1}{{}^{2k+1}{C}_1}+....+\frac{1}{{}^{2k+1}{C}_{2k}}-\frac{1}{{}^{2k+1}{C}_{2k+1}}=0$