xét hàm $f(x)=x+x^2+x^3+...+x^{2010}=x(1+x+x^2+...+x^{2009})=x.\frac{x^{2010}-1}{x-1}$ Ta có $f'(x)=1+2x+3x^2+....+2010.x^{2009}$Nhưng ta cũng có $f'(x)={\frac {2011\,{x}^{2010}-1}{x-1}}-{\frac {x\cdot ({x}^{2010}-1)}{ \left( x-1 \right) ^{2}}}$Vậy $D=f'(3)=\frac{2011.3^{2010}-1}{2}-\frac{3.(3^{2010}-1)}{4}$
xét hàm $f(x)=x+x^2+x^3+...+x^{2010}=x(1+x+x^2+...+x^{2009})=x.\frac{x^{2010}-1}{x-1}$ Ta có $f'(x)=1+2x+3x^2+....+2010.x^{2009}$Nhưng ta cũng có $f'(x)={\frac {2011\,{x}^{2010}-1}{x-1}}-{\frac {x\cdot ({x}^{2010}-1)}{ \left( x-1 \right) ^{2}}}$Vậy $D=f'(3)=\frac{2011.3^{2010}-1}{2}-\frac{3.(3^{2011}-1)}{4}$
xét hàm $f(x)=x+x^2+x^3+...+x^{2010}=x(1+x+x^2+...+x^{2009})=x.\frac{x^{2010}-1}{x-1}$ Ta có $f'(x)=1+2x+3x^2+....+2010.x^{2009}$Nhưng ta cũng có $f'(x)={\frac {2011\,{x}^{2010}-1}{x-1}}-{\frac {x\cdot ({x}^{2010}-1)}{ \left( x-1 \right) ^{2}}}$Vậy $D=f'(3)=\frac{2011.3^{2010}-1}{2}-\frac{3.(3^{201
0}-1)}{4}$