Given, f(x)=\sum_\limits{k=1}^{10} k x^{k} $\begin{aligned} & f(x)=x+2 x^2+\ldots \ldots \ldots+10 x^{10} \\\\ & f(x) . x=x^2+2 x^3+\ldots \ldots \ldots+9 x^{10}+10 x^{11} \\\\ & f(x)(1-x)=x+x^2+x^3+\ldots \ldots \ldots+x^{10}-10 x^{11} \\\\ & \therefore f(x)=\frac{x\left(1-x^{10}\right)}{(1-x)^2}-\frac{10 x^{11}}{(1-x)} \end{aligned}$ $\Rightarrow$ $f(x)=\frac{x-x^{11}-10 x^{11}+10 x^{12}}{(1-x)^2} = \frac{10 x^{12}-11 x^{11}+x}{(1-x)^2}$ So, $\begin{aligned} & f(2)=2\left(1-2^{10}\right)+10 \cdot 2^{11} \\\\ & =2+18 \cdot 2^{10} \end{aligned}$ $f'\left( x \right) = {{{{\left( {1 - x} \right)}^2}\left( {120{x^{11}} - 121{x^{10}} + 1} \right) + 2\left( {1 - x} \right)\left( {10{x^{12}} - {{11.x}^{11}} + 2} \right)} \over {{{\left( {1 - x} \right)}^4}}}$ So, $f'\left( x \right) = {{1\left( {{{120.2}^{11}} - {{121.2}^{10}} + 1} \right) + 2\left( { - 1} \right)\left( {{{10.2}^{12}} - {{11.2}^{11}} + 2} \right)} \over {{{\left( { - 1} \right)}^4}}}$ = ${2^{10}}\left( {83} \right) - 3$ Hence $2 f(2)+f^{\prime}(2)=119.2^{10}+1$ $\Rightarrow \text { So, } \mathrm{n}=10$