Question

Find the greatest term of the following expansion :
$(2x+5y{)}^{12}$ when $x=8$ and $y=3$

Answer 1

If $T_{r+1}$ be the greatest term in the expansion of $(X+Y{)}^n$ then

$\left|\frac{T_{r+1}}{T_r}\right|\ge 1$

or, $|\frac{n-r+1}{r}.\frac{Y}{X}|\ge 1$

Here, $X=2x,$ $Y=5y,$ $n=12$ and $x=8$ and $y=3$

or, $|\frac{12-r+1}{r}.\frac{5y}{2x}|\ge 1$

or, $|\frac{(13-r)}{r}.\frac{15}{16}|\ge 1$

or, $\frac{195-15r}{16r}\ge 1$

or, $195-15r\ge 16r$

or, $31r\le 195$

or, $r\le \frac{195}{31}$

$\therefore r=\left[\frac{195}{31}\right]=6$

$\therefore$ Greatest term is $T_7$ or $7th.$ $[\because T_{r+1}$ is the greatest term$]$