8.0 Numerically Greatest term

If ${T_r}$ and ${T_{r + 1}}$ be the ${r^{th}}$ and ${(r + 1)^{th}}$ terms in the expansion of ${(a + x)^n}$, then

$$\left( {\frac{{{T_{r + 1}}}}{{{T_r}}}} \right) = \frac{{^n{C_r}{a^{n - r}}{x^r}}}{{^n{C_{r - 1}}{a^{n - r + 1}}{x^{r - 1}}}} = \left| {\frac{{n - r + 1}}{r}} \right|\left| {\frac{x}{a}} \right| = \left| {\frac{{n + 1}}{r} - 1} \right|\left| {\frac{x}{a}} \right|$$

Let numerically, ${T_{r + 1}}$ be the greatest term in the above expansion. Then,

$$\begin{equation} \begin{aligned} {T_{r + 1}} \geqslant {T_r} \\ \frac{{{T_{r + 1}}}}{{{T_r}}} \geqslant 1 \\ \left| {\frac{{n + 1}}{r} - 1} \right|\left| {\frac{x}{a}} \right| \geqslant 1 \\ \frac{{n + 1}}{r} \geqslant 1 + \left| {\frac{a}{x}} \right| \\ \frac{{n + 1}}{{1 + \left| {\frac{a}{x}} \right|}} \geqslant r \\\end{aligned} \end{equation} $$

From the above expression, calculate the value of $r$ keeping in mind that $r$ must be an integer and then find the value of ${T_{r + 1}}$ which will be the numerically greatest term.