Question

In the following expansion find which is the greatest term:
${(2x-3y)}^{28}$, when $x=9,y=4$.

Answer 1

General term in the binomial expansion of ${(a+b)}^n$ is $T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$ and
$T_{r+1}>T_r$

i.e. ${}^n{C}_r\cdot a^{n-r}\cdot b^r>{}^n{C}_{r-1}\cdot a^{n-r+1}\cdot b^{r-1}$

$\Rightarrow \frac{n-r+1}{r}\times b>a$

$\Rightarrow \frac{n-r+1}{r}>\frac{a}{b}$

$\Rightarrow n-r+1>\frac{ra}{b}$

$\Rightarrow n+1>r(1+\frac{a}{b})$

$\Rightarrow r<b\left(\frac{n+1}{a+b}\right)$

$\Rightarrow r<12\left(\frac{29}{30}\right)=11.6$

$\therefore r=11$

Thus, $T_{12}$ is greatest.

$T_{12}={}^{29}{C}_{11}{\left(18\right)}^{17}{(-12)}^{11}$