Question

If $S_n$ = $\sum _{r=0}^n\frac{1}{{}^n{C}_r}$ and $t_n$ = $\sum _{r=0}^n\frac{r}{{}^n{C}_r}$, then $\frac{t_n}{s_n}$, when n =100 equals to

__

Answer 1

We will try to express $t_n$ in terms of $s_n$.

$t_n$ = $\sum _{r=0}^n\frac{r}{{}^n{C}_r}$ and $S_n$ = $\sum _{r=0}^n\frac{1}{{}^n{C}_r}$

We can write $\frac{r}{{}^n{C}_r}$ as $\frac{(r+1)-1}{{}^n{C}_r}$ or $\frac{n-(n-1)}{{}^n{C}_r}$. If we write it as $\frac{(r+1)-1}{{}^n{C}_r}$, then we will get the sum in terms of

$t_n$+1 or we may not be able to proceed. So we will go with $\frac{n-(n-1)}{{}^n{C}_r}$

⇒ $t_n$ = $\sum _{r=0}^n\frac{n-(n-1)}{{}^n{C}_r}$

= $\sum _{r=0}^n\frac{1}{{}^n{C}_r}-\sum _{r=0}^n\frac{n-r}{{}^n{C}_r}$

[because ${}^n{C}_r{=}^n{C}_{n-r}$]

= n$\sum _{r=0}^n\frac{1}{{}^n{C}_r}-\sum _{u=n}^0\frac{u}{{}^n{C}_u}$

[we replace n-r by u. The boundaries also changes. When r = 0, u = n and when r = n, u = 0]

⇒ $t_n$ = n $\times$ $s_n$ - $\sum _{u=n}^n\frac{u}{{}^n{C}_u}$

= n $s_n$ - $\sum _{r=0}^n\frac{r}{{}^n{C}_u}$

[We replaced u by r. Change of variable does not affect the sum]

⇒ $t_n$ = n $s_n$ = $t_n$

⇒2 $t_n$ = n $s_n$

$\frac{t_n}{s_n}$ = $\frac{n}{2}$

When n = 100 $\frac{t_n}{s_n}$ = $\frac{100}{2}$ = 50