Find the index 'n' of binomial $(\frac{x}{5}+\frac{2}{5})^n$ if the 9th term has numerically the greatest coefficient
My working:
For $T_{r+2}>T_{r+1}$
$\frac{\binom{n}{r+1}(2)^{n-r-1}}{5^n}>\frac{\binom{n}{r}(2)^{n-r}}{5^n}$
This simplifies to
$\frac{n-r}{r+1}>2$
$n>3r+2$ or $r<\frac{n-2}{3}$----(1)
Similarly for $T_{r+2}
We have to find R such that R-1 satisfies (1) and R+1 satisfies (2)
I think I've almost reached the solution but I can't solve further. I need some hint now.
We begin by factoring out $1/5$ from the binomial.
$$\left(\frac{x}{5}+\frac{2}{5}\right)^n=\left(\frac{1}{5}\right)^n(x+2)^n$$
We now seek the integer exponent $n$ such that the coefficient of the 9th term of $(x+2)^n$ is maximum.
Since we seek the coefficients of $(x+2)^n$, we can simpify our work by allowing $x=1$. So we work with $(1+2)^2$.
By the binomial theorem, we know that $$(1+2)^n=\sum_{r=0}^n \binom{n}{r}1^{n-r}\cdot2^r=\sum_{r=0}^n \binom{n}{r}2^r\ .$$
We need to calculate specific terms, so let $T_r=\binom{n}{r}2^r$.
Notice that the first term of the expansion is labeled $r=0$. We seek that 9th term, so $r=8$.
In order for $T_8$ to be the maximum coefficient, we need that $T_8>T_7$ and that $T_8>T_9$.
Let's find the condition for $T_8>T_7$.
\begin{align} T_8&>T_7\\ \binom{n}{8}2^8&>\binom{n}{7}2^7\\ \frac{n!}{8!(n-8)!}2^8&>\frac{n!}{7!(n-7)!}2^7\\ (n-7)\cdot2&>8\\ n&>11 \end{align}
Now the condition so that $T_8>T_9$.
\begin{align} T_8&>T_9\\ \binom{n}{8}2^8&>\binom{n}{9}2^9\\ \frac{n!}{8!(n-8)!}2^8&>\frac{n!}{9!(n-9)!}2^9\\ 9&>(n-8)\cdot2\\ n&<12.5 \end{align}
So, we need an integer $n$ such that $11
Hint: We have that $$T_9 > T_8 \Rightarrow \binom {n}{8}2^{n-8} > \binom {n}{7}2^{n-7} \Rightarrow n> 23$$
Also, $$T_{10}< T_9 \Rightarrow?$$
Hope you can take it from here.