Let f(x) = ∑ _ k = 1 ^ 10 k · x ^ k , x ∈ ℝ , if 2f(2) + f ^ ' (2)...

Given,

$f\left(x\right) = \mathop{∑}_{k = 1}^{10}k \cdot {x}^{k}, x \in \mathbb{R}$

$⇒ f\left(x\right) = x + 2{x}^{2} + 3{x}^{3} + \ldots \ldots \ldots + 10{x}^{10}$

Now let $S = x + 2{x}^{2} + 3{x}^{3} + \ldots \ldots \ldots + 10{x}^{10}$

So, $S \cdot x = {x}^{2} + 2{x}^{3} + \ldots \ldots \ldots \ldots + 9{x}^{10} + 10{x}^{11}$

Now subtracting above equations we get,

$S\left(1 - x\right) = x + {x}^{2} + {x}^{3} + \ldots \ldots + {x}^{10} - 10{x}^{11}$

$⇒ S\left(1 - x\right) = \dfrac{x\left(1 - {x}^{10}\right)}{1 - x} - 10{x}^{11}$

$⇒ S = \dfrac{x\left(1 - {x}^{10}\right)}{{\left(1 - x\right)}^{2}} - \dfrac{10{x}^{11}}{\left(1 - x\right)}$ = f$\left(x\right)$

Now finding, $f\left(2\right) = 2\left(1 - {2}^{10}\right) + 10 \cdot {2}^{11}$

$⇒ f\left(2\right) = 2 + 18 \cdot {2}^{10}$

Now finding ${f}^{'}\left(x\right)$ we get,

${f}^{'}\left(x\right) = \dfrac{- 10{x}^{11}}{{\left(1 - x\right)}^{2}} - \dfrac{110{x}^{10}}{1 - x} - \dfrac{10{x}^{10}}{{\left(1 - x\right)}^{2}} + \dfrac{2x\left(1 - {x}^{10}\right)}{{\left(1 - x\right)}^{3}} + \dfrac{1 - {x}^{10}}{{\left(1 - x\right)}^{2}}$

$⇒ {f}^{'}\left(2\right) = \dfrac{- 10 \cdot {2}^{11}}{{\left(1 - 2\right)}^{2}} - \dfrac{110 \cdot {2}^{10}}{1 - 2} - \dfrac{10 \cdot {2}^{10}}{{\left(1 - 2\right)}^{2}} + \dfrac{2 \cdot 2 \cdot \left(1 - {2}^{10}\right)}{{\left(1 - 2\right)}^{3}} + \dfrac{1 - {2}^{10}}{{\left(1 - 2\right)}^{2}}$

$⇒ {f}^{'}\left(2\right) = - 10 \cdot {2}^{11} + 110 \cdot {2}^{10} - 10 \cdot {2}^{10} - 4\left(1 - {2}^{10}\right) + 1 - {2}^{10}$

$⇒ {f}^{'}\left(2\right) = 83 \cdot {2}^{10} - 3$

$2f ( 2 ) + {f}^{'} ( 2 ) = 2\left(2 + 18 \cdot {2}^{10}\right) + 83 \cdot {2}^{10} - 3 = 119 ( 2{)}^{10} + 1$

$∴ n = 10$