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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 $D=f'(3)=\frac{2011.3^{2010}-1}{2}-\frac{3.(3^{2010}-1)}{4}$
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