`(1+2x)^{n}=a_{0}+a_{1}x^{1}+a_{2}x^{2}+...+a_{n}x^{n}` and `a_{0}+(a_1)/(2)+(a_2)/(2^2)+...+(a_n)/(2^n)=4096`

Find maximum coefficient in the epansion `(1+2x)^n`

With x = $\frac{1}{2}$ on expression 1: $a_{0}$ + $a_{1}$$x^{1}$ +,... $a_{n}$x^n

=> $(1+2. $\frac{1}{2}$ ) ^{n}$  = 4096

=> n = 12

So we have the expression :$ ( 1 + 2x )^{12}$

General class number : $T_{k}$ = 12(k)k . $2^{k}$ .$x^{k}$ 

=> the max coefficient : u = 12C(k) . $2^{k}$ ( 0 $\leq$ k $\leq$ 12 )

to get the max coefficient => k max => k - 1 $\leq$ k $\leq$ k+1 ( because it must greater than the last and front number )

So we have 2 new inequality :

12C(k-1) . $2^{k-1}$ < 12C(k) . $2^{k}$ (a)

12C(k+1) . $2^{k-1}$ < 12C(k) . $2^{k}$ (b)

    *Solve upper inequality, we get: *

k < $\frac{26}{3}$ 

k > $\frac{23}{3}$ 

And k ∈ Z

=> k = 8

=> The max coefficient : 12C8 . $2^{8}$ = 126720