Xét $f(x) = -(1-x)^{2n+1}$
$f(x) = -\left(\sum_{i=0}^{2n+1}C_{2n+1}^{i}(-1)^ix^i\right)$
$f'(x) = -\left(\sum_{i=1}^{2n+1}C_{2n+1}^{i}(-1)^iix^{i-1}\right)$
$f'(x) = 1.x^0C_{2n+1}^1-2.x^1C_{2n+1}^2+3x^2C_{2n+1}^3 +....+(-1)^iix^{i-1}C_{2n+1}^i+...+(2n+1)x^{2n}C_{2n+1}^{2n+1}$
thay x = 2 ta được
$f'(2)=1.2^0C_{2n+1}^1-2.2^1C_{2n+1}^2+32^2C_{2n+1}^3 +....+(-1)^iix^{i-1}C_{2n+1}^i+...+(2n+1)2^{2n}C_{2n+1}^{2n+1}$
Mặt khác $f'(x) = (2n+1)(1-x)^{2n}$
nên $f'(2) = (2n+1)(1-2)^{2n} = (2n+1)$
Vậy $1.2^0C_{2n+1}^1-2.2^1C_{2n+1}^2+32^2C_{2n+1}^3 +....+(-1)^iix^{i-1}C_{2n+1}^i+...+(2n+1)2^{2n}C_{2n+1}^{2n+1} = (2n+1) =2005$
suy ra $n =1002$
Nhớ vote nhé