Dùng công thức $C_n^k=\frac{k+1}{n+1}.C_{n+1}^{k+1}$ (tự chứng minh nhé)Ta có $C_n^0=\frac 1{n+1}C_{n+1}^1; C_n^2=\frac 3{n+1}C_{n+1}^3...;C_n^n=C_{n+1}^{n+1}$
Suy ra $VT=\frac{1}{n+1}.2C_{n+1}^1+\frac 1{n+1}2^2C_{n+1}^2+...+\frac 1{n+1}2^{n+1}C_{n+1}^{n+1}+1$
Rút gọn 2 vế cho $\frac 1{n+1}$
Ta đc $C_{n+1}^0+2C_{n+1}^1+2^2C_{n+1}^2+...2^{n+1}C_{n+1}^{n+1}=3^{2017}$
$\Leftrightarrow (1+2)^{n+1}=3^{2017}$
$\Leftrightarrow n=2016$