Question

if the second term of the expression[a^1/13+a/√(a^-1)]ⁿ is 14a^​5/2 then find the value of ⁿC₃/ⁿC₂

Answer 1

$$\begin{gathered} {\left[{\mathrm{a}}^{\frac{1}{13}}+\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^{-1}}}\right]}^{\mathrm{n}} \\ {\mathrm{T}}_{\mathrm{r}+1}={}^{\mathrm{n}}\mathrm{c}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{n}-\mathrm{r}}{\mathrm{y}}^{\mathrm{r}} \\ \mathrm{r}+1=2 \\ \mathrm{r}=1. \\ {\mathrm{T}}_2={}^{\mathrm{n}}\mathrm{c}_1{\left({\mathrm{a}}^{\frac{1}{13}}\right)}^{\mathrm{n}-1}{\left(\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^{-1}}}\right)}^1 \\ ={}^{\mathrm{n}}\mathrm{c}_1\left({\mathrm{a}}^{\frac{\mathrm{n}-1}{13}}\right)\left({\mathrm{a}}^{\frac{3}{2}}\right) \\ ={}^{\mathrm{n}}\mathrm{c}_1\left({\mathrm{a}}^{\frac{2\mathrm{n}-2+39}{26}}\right) \\ ={}^{\mathrm{n}}\mathrm{c}_1{\mathrm{a}}^{\frac{2\mathrm{n}+37}{26}}=14{\mathrm{a}}^{\frac{5}{2}} \\ \mathrm{So}, \\ \frac{2\mathrm{n}+37}{26}=\frac{5}{2} \\ 2\mathrm{n}+37=65 \\ 2\mathrm{n}=28 \\ \mathrm{n}=14. \\ \frac{{}^{\mathrm{n}}\mathrm{c}_3}{{}^{\mathrm{n}}\mathrm{c}_2}=\frac{{}^{14}\mathrm{c}_3}{{}^{14}\mathrm{c}_2} \\ =\frac{{\displaystyle \frac{14!}{11!}}}{{\displaystyle \frac{14!}{12!}}}=\frac{12!}{11!}=12\mathrm{ANS}... \end{gathered}$$