Greatest term in the expansion $(2+3x)^9$ I have got a question which is stated as

Find numerically the greatest term in the expansion of $(2+3x)^9$ where $x=\frac{3}{2}$

I haven't any idea how to proceed but I just plugged in value of $x$ and got $$(2+3x)^9=\bigg(\frac{13}{2}\bigg)^9$$
How can I determine greatest term here?
Please help!!!
 A: The term involving $(3/2)^n$ is
$$ a_n = {9 \choose n} 2^{9-n} \left( {9 \over 2} \right)^n$$
and so you have
$$ {a_n \over a_{n-1}} = {{9 \choose n} \over {9 \choose n-1}} {2^{9-n} \over 2^{9-(n-1)}} {(9/2)^n \over (9/2)^{n-1}} $$
or after some simplification
$$ {a_n \over a_{n-1}} = {{9 \choose n} \over {9 \choose n-1}} \times {9 \over 4}. $$
Now attack that quotient of binomial coefficients; it's
$$ {{9 \choose n} \over {9 \choose n-1}} = {9! \over n! (9-n)!} {(n-1)! (10-n)! \over 9!} = {(n-1)! \over n!} {(10-n)! \over (9-n)!} = {10 - n \over n}.$$
So you have 
$$ {a_n \over a_{n-1}} = {10 - n \over n} \times {9 \over 4} $$
and this decreases as $n$ increases.  So $a_n/a_{n-1} > 1$ for small $n$ and $a_n/a_{n-1} < 1$ for large $n$ - the sequence of $a_n$ increases and then decreases as $n$ increases.  Find the $n$ where this changeover happens.
A: You have 
$$(3 x+2)^9=19683 x^9+118098 x^8+314928 x^7+489888 x^6+489888 x^5+326592 x^4+145152 x^3+41472 x^2+6912 x+512$$
Plugging $x=1.5$ the terms, from degree $0$ up are
$$512,10368,93312,489888,1.65337\times 10^6,3.72009\times 10^6,5.58013\times 10^6,5.38084\times 10^6,3.02672\times 10^6,756681$$
the largest terms is  for $x^6$ and is
$5.58013\times 10^6$
A: HINT:
The middle term of the expansion is the greatest. For odd $n$, the greatest term will be for $r=\frac{n+1}{2}$ and for even $n$, there will be two greatest terms and they will be for $r=\frac{n}{2}$ and $r=\frac{n}{2}+1$.
Hope this helps.
