Question

Find the power $n$ of the binomial ${(\frac{x}{5}+\frac{2}{5})}^n$, if the ninth term of the expansion has the greatest coefficient.

Answer 1

Given:

${(\frac{x}{5}+\frac{2}{5})}^n$

General term of given expansion is

$T_{r+1}={\sum }_{r=0}^n{}^n{C}_r{\left(\frac{x}{5}\right)}^{n-r}{\left(\frac{2}{5}\right)}^r$

$9^{th}$ term When $r=8$

$T_{8+1}={}^n{C}_8{\left(\frac{x}{5}\right)}^{n-8}{\left(\frac{2}{5}\right)}^8$

$T_9={}^n{C}_8{x}^{n-8}{2}^8{\left(\frac{1}{5}\right)}^{8+n-8}$

$T_9={}^n{C}_8{x}^{n-8}{2}^8\left(\frac{1}{5^n}\right)$

As we know that in the binomial expansion the middle term has the greatest value

So if n is even ,then $\frac{n}{2}=9\Rightarrow n=18$

Hence

$T_9={}^{18}{C}_8{x}^{10}{2}^8\left(\frac{1}{5^{18}}\right)$

Coefficient of above term is ${}^{18}{C}_8{2}^8\left(\frac{1}{5^{18}}\right)$ ..... $\left(1\right)$

When n is odd then $9^{th}$ term can be middle term of $\frac{n+1}{2}=9$ and $\frac{n+3}{2}=9$

$9^{th}$ term can be middle term of $n=17$ and $n=15$

When $n=17$

$T_9={}^{17}{C}_8{x}^9{2}^8\left(\frac{1}{5^{17}}\right)$

Coefficient ${}^{17}{C}_8{2}^8\left(\frac{1}{5^{17}}\right)$ ...... $\left(2\right)$

When $n=15$

$T_9={}^{15}{C}_8{x}^7{2}^8\left(\frac{1}{5^{15}}\right)$

Coefficient ${}^{15}{C}_8{2}^8\left(\frac{1}{5^{15}}\right)$ ..... $\left(3\right)$

Comparing equations $\left(2\right)$ and $\left(3\right),$ we get

${}^{15}{C}_8{2}^8\left(\frac{1}{5^{15}}\right)$ is greatest

But we should also compare eq (3) and (1) ,we get

${}^{15}{C}_8{2}^8\left(\frac{1}{5^{15}}\right)$ is greatest

Hence $n=15$