Question

Find the numerically greatest term in the expansion of $(2x+y{)}^{14}$ when $x=4$and $y=5$.

Answer 1

$\begin{array}{c}{T}_{r+1}\ge T_n\\ {}^n{C}_r{\left(2x\right)}^r{y}^{n-r}{\ge }^n{C}_{r-1}{\left(2x\right)}^{r-1}{y}^{n-r+1}\\ \frac{{}^n{C}_r}{{}^n{C}_{r-1}}\ge \frac{y}{2x}\\ \frac{{}^{14}{C}_r}{{}^{14}{C}_{r-1}}\ge \frac{5}{8}\\ \frac{\frac{14!}{r!\left(14-r\right)!}}{\frac{14!}{\left(r-1\right)!\left(14-r+1\right)!}}\ge \frac{5}{8}\\ \frac{14-r+1}{r-1}\ge \frac{5}{8}\\ 8(14-r+1)\ge 5(r-1)\\ 112-8r+8\ge 5r-1\\ 13r\le 125\\ r\le \frac{125}{13}\\ r_{max}=9\end{array}$

So, the greatest term is $T_{10}$