cho (2x+1)^n=a0+a1.x+a2.x^2+...+an.x^n thỏa mãn a0+a1/2+a2/2^2+...+an/2^n

Câu hỏi :

Cho ${\left(2x+1\right)}^n=a_0+a_{1}x+a_2{x}^2+...+a_n{x}^n$ thỏa mãn $a_0+\frac{a_1}{2}+\frac{a_2}{2^2}+...+\frac{a_n}{2^n}=4096$ Tìm $a_5$