Correct Answer - Option 3 :
\(\frac{{n\,{a_n}}}{2}\)
Concept:
Combination:
It's a mathematical tool that determines the number of possible arrangements in a collection of items. Where the order of selection doesn't matter.
\(C(n, r) = {^nc_r} = \dfrac{n!}{(n-r)!r!}\)
n : number of items in set
r : number of items selected from the set.
Calculation:
Given \(a_n = \displaystyle\sum_{r=0}^n \dfrac{1}{^nc_r}\)
\(a_n = \displaystyle\sum_{r=0}^n \dfrac{r! (n-r)!}{n!}\)
\(a_n = \dfrac{0!n!}{n!} + \dfrac{1! (n-1)!}{n!} + \dfrac{2!(n-2)!}{n!}+ ... + \dfrac{n!(n-n)!}{n!}\)
\(=1 + \dfrac{1!(n-1)!}{n(n-1)!} + \dfrac{2! (n-2)!}{n(n-1)(n-2)!} + ...+ 1\)
\(=2 + \dfrac{1}{n} + \dfrac{2}{n(n-1)} + \dfrac{3 \times 2}{n(n-1)(n-2)} + ...\) ---(1)
Now let \(x = \displaystyle\sum_{r=0}^n \dfrac{r \times r! (n-r)!}{n!}\)
\(=\displaystyle\sum_{r=0} \left[\dfrac{0\times 0! (n!)}{n!} + \dfrac{1 \times 1! (n-1)!}{n!} + \dfrac{2 \times 2! (n-2)!}{n!}+... \right]\)
\(=0 + \dfrac{1!}{n} + 2 \cdot \dfrac{2!}{n(n-1)} + 3 \dfrac{3!}{n(n-1)(n-2)}+... + n \dfrac{n!(n-n)!}{n!}\) ---(2)
\(x = \displaystyle\sum_{r=0}^n \dfrac{n-(n-r)}{^nc_r}\)
\(=\displaystyle\sum_{r=0}^n \dfrac{n}{^nc_r} - \sum_{r=0}^n \dfrac{r}{^nc_r}\)
We know that \({^nc_r} = {^nc_{n-r}}\)
\(= n \displaystyle\sum_{r=0}^n \dfrac{1}{^nc_r} - \sum \dfrac{n-r}{^nc_{n-r}}\)
\(= n \cdot a_n - \displaystyle\sum_{r=0}^n \dfrac{n-r}{^nc_{n-r}}\)
Given
\(x = \displaystyle\sum_{r=0}^n \dfrac{r}{^nc_r}\)
replace r with n - r
\(x = \displaystyle\sum_{r=0}^n \dfrac{n-r}{^nc_{n-r}}\)
x = n an - 3x
2x = nan
\(x = \dfrac{n}{2}a_n\)
option 'c' is correct.
