Binomial theorem for positive integral Index
Question

# For n$\ge$2, let  and Column-IColumn-IIA. $\sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{\mathrm{r}}{{\mathrm{C}}_{\mathrm{r}}}$P. ${\mathrm{na}}_{\mathrm{n}}$B. $\sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{\mathrm{n}-\mathrm{r}}{{\mathrm{C}}_{\mathrm{r}}}$Q. $\frac{1}{\mathrm{n}}{\mathrm{a}}_{\mathrm{n}-1}$C. R. $\frac{1}{2}{\mathrm{na}}_{\mathrm{n}}$D.  $\sum _{\mathrm{r}=0}^{\mathrm{n}-1} \frac{1}{\left(\mathrm{n}-\mathrm{r}\right){\mathrm{C}}_{\mathrm{r}}}$S. $\frac{1}{2\mathrm{n}}{\mathrm{a}}_{\mathrm{n}-1}$

Moderate
Solution

## using $⇒2\sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{\mathrm{r}}{{\mathrm{C}}_{\mathrm{r}}}={\mathrm{na}}_{\mathrm{n}}⇒\sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{\mathrm{r}}{{\mathrm{C}}_{\mathrm{r}}}=\frac{1}{2}{\mathrm{na}}_{\mathrm{n}}$similarly, $\sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{\mathrm{n}-\mathrm{r}}{{\mathrm{C}}_{\mathrm{r}}}=\frac{1}{2}{\mathrm{na}}_{\mathrm{n}}$using, ${\mathrm{rC}}_{\mathrm{r}}={\mathrm{nC}}_{\mathrm{r}-1}$ we getsimilarly ,$\sum _{\mathrm{r}=0}^{\mathrm{n}-1} \frac{1}{\left(\mathrm{n}-\mathrm{r}\right){\mathrm{C}}_{\mathrm{r}}}=\frac{1}{\mathrm{n}}{\mathrm{a}}_{\mathrm{n}-1}$

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