Mathematics - Binomial Theorem Question with Solution | TestHub

Note that ${\mathrm{b}}_{\mathrm{n}}={\sum }_{\mathrm{r}=0}^{\mathrm{n}}\frac{\mathrm{r}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}={\sum }_{\mathrm{r}=0}^{\mathrm{n}}\frac{\mathrm{n}-\mathrm{r}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}={\mathrm{na}}_{\mathrm{n}}-{\mathrm{b}}_{\mathrm{n}}(\mathrm{n}\ge 0)$ So that ${\mathrm{na}}_{\mathrm{n}}=2{\mathrm{b}}_{\mathrm{n}}$ for $\mathrm{n}\ge 0$ and hence it follows that ${\mathrm{c}}_{\mathrm{n}}=\frac{2}{\mathrm{n}}$ for all $\mathrm{n}\ge 1$. Thus we want to find ordered pairs $(p,q)$ of positive integers such that $2p^{-1}+2q^{-1}=1$ or $2(\mathrm{p}+\mathrm{q})=\mathrm{pq}$ or $(\mathrm{p}-2)(\mathrm{q}-2)=4$ Thus $\mathrm{p}-2$ can take the values $1,2,4$ only and there are exactly 3 ordered pairs with the desired property, $(3,6),(4,4)$ and $(6,3)$.