Question

If ${\sum }_{r=0}^n\frac{r}{{}^n{C}_r}={\sum }_{r=0}^n\frac{n^2-3n+3}{{2.}^n{C}_r},$ then

• A. n = 1
• B. n = 2
• C. n = 3
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A

n = 1

C

n = 3

${\sum }_{r=0}^n\frac{r}{{}^n{C}_r}={\sum }_{r=0}^n\frac{n-r}{{}^n{C}_{n-r}}=\frac{n}{2}{\sum }_{r=0}^n\frac{1}{{}^n{C}_r}\therefore \frac{n}{2}=\frac{n^2-3n+3}{2}\Rightarrow n=1,3$