If the sum of the coefficients in the expansion of  $(1+2x{)}^n$is $6561$, then the greatest coefficient in the expansion, is 

Sum of the coefficients in the given expansion=f(1)=$3^n=6561=3^8$ then $n= 8.$

We have to find the greatest coefficients. This means that we 

have to determine the greatest term for$x = 1$.

$\therefore \frac{T_{r+1}}{T_r}=\frac{9-r}{r}\times 2x$                      [See Ex. 16] 

$\Rightarrow \frac{T_{r+1}}{T_r}=\frac{(9-r)\times 2}{r}$                          $[\because x=1]$

Now  $\frac{T_{r+1}}{T_r}>1\Rightarrow 18-2r>r\Rightarrow r<6$  

Thus, $6th$ and $7th$terms are greatest and are equal in

magnitude. 

Now,  

$\begin{array}{l}{T}_6{=}^8{C}_5\left(1{)}^{8-5}\right(2{)}^5[\because x=1]\\ \Rightarrow T_6{=}^8{C}_5\times 2^5=1792\end{array}$