Question

The index 'n' of the binomial ${(\frac{x}{5}+\frac{2}{5})}^n$ if the $9$th term of the expansion has greatest the coefficient $(n\in N)$ is :

Answer 1

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

$\frac{T_{r+1}}{T_r}=\frac{{}^n{C}_{r+1}{2}^{n-r-1}x}{{}^n{C}_r{2}^{n-r}}=\frac{n-r}{2(r+1)}$

Case 1: $r=8$ (Starting from $0^{th}$ power of x)

ie, $\frac{T_9}{T_8}=\frac{n-8}{18}$

as, $T_9$ is the greatest,

the smallest $n$ which satisfies the inequality $\frac{T_9}{T_8}\ge 1$ is the index of the binomial.

ie, $n-8\ge 18\Rightarrow n\ge 26$

So, $n=26$

Case 2: $r=n-8$ (Starting from $n^{th}$ power of x)

ie, $\frac{T_{n-7}}{T_{n-8}}=\frac{8}{2(n-7)}$

as, $T_{n-7}$ is the greatest,

the greatest $n$ which satisfies the inequality $\frac{T_{n-7}}{T_{n-8}}\ge 1$ is the index of the binomial.

ie, $8\ge 2n-14\Rightarrow n\le 11$

So, $n=11$