Sum of coefficients of a binomial expansion Here I have a question:
If the sum of the coefficients in the expansion of $(1+2x)^n $ is $ 6561$ then the greatest term in the expansion for $x=1/2 $ is? 
So I used the formula for greatest term in an expansion and got:
$$ r < \frac {n+1} {2} $$
But I don't understand how to use the value of sum of coefficients in the problem. I don't know any formula for that. Also in the solutions I saw them taking $ x=1 $ and directly equating with $6561$ and getting $n=8$. What's the logic behind? I can't seem to understand. 
 A: Hint The sum $a_n + \cdots + a_0$ of the coefficients of a polynomial $p(x) := a_n x^n + \cdots + a_1 x + a_0$ coincides with $p(1)$.
A: Note that
\begin{align*}
(1 + 2 x)^n &= \sum_{k=0}^n \binom{n}{k} 1^{n-k} (2x)^{k} \\
& = \sum_{k=0}^n  2^k \binom{n}{k}   x^{k} \\
& = \sum_{k=0}^n a_k x^k,
\end{align*}
where $a_k = 2^k \binom{n}{k}$. Then, the sum of the coefficients is:
\begin{align*}
\sum_{k=0}^n a_k & =  \sum_{k=0}^n a_k 1^k \\
& = (1 + 2 )^n \\
& = 3^n
\end{align*}
where we used the special case $x = 1$. To find the value of $n$ for which $3^n = 6561$, we can take logarithms and note:
$$ \log 3^n = n \log 3 = \log 6561$$
which implies $n = \log 6561 / \log 3 = 8$.
Therefore we have for $x = 1/2$
$$ (1 + 2 x)^n = 2^n = 2^8 = 256.$$
A: The sum of coefficient in a polynomial is found by evaluating the polynomial at $x=1$.You have already found $n=8$ by substituting $x=1$ in $(1+2x)^n = 6561.$ 
Your question is what is the greatest term in the expansion of $(1+2x)^n$ at $x=1/2.$ 
With $n=8$ and $x=1/2$ we have $(1+2x)^n =2^8=256$ which is a constant.Thus the greatest term is 256.   
