xét hàm f(x)=x+x2+x3+...+x2010=x(1+x+x2+...+x2009)=x.x2010−1x−1 Ta có f′(x)=1+2x+3x2+....+2010.x2009Nhưng ta cũng có f′(x)=2011x2010−1x−1−x⋅(x2010−1)(x−1)2Vậy $D=f'(3)=\frac{2011.3^{2010}-1}{2}-\frac{3.(3^{2011}-1)}{4}$
xét hàm
f(x)=x+x2+x3+...+x2010=x(1+x+x2+...+x2009)=x.x2010−1x−1 Ta có
f′(x)=1+2x+3x2+....+2010.x2009Nhưng ta cũng có
f′(x)=2011x2010−1x−1−x⋅(x2010−1)(x−1)2Vậy $D=f'(3)=\frac{2011.3^{2010}-1}{2}-\frac{3.(3^{201
0}-1)}{4}$